Raise the quality in parentheses to the indicated exponent, and simplify the resulting expression with positive exponents.

Maximum is the highest a graph can reach. In this case the graph continues forever therefore the maximum is:

infinity or ∞

The minimum is the lowest place the graph reaches. In this case it would be:

-6

The zeros are where the graph intersects the x axis. In this case it would have two zeros, which are:

(-3, 0) and (0.5, 0)

Hope this helped!

~Just a girl in love with Shawn Mendes

c. Reflection over either the x-axis or y-axis will change the graph

This is false because [tex]y=sin(x)[/tex] is an odd function, not an even one. This means that [tex]sin(-x)=-sin(x)[/tex], and a reflection over the y-axis will change the graph.

This is false because we said that [tex]sin(-x)=-sin(x)[/tex]

This is true. Since [tex]sin(x)[/tex] is an odd function, then reflection over either the x-axis or y-axis will change the graph as we said in a. So, for [tex]f(x)[/tex]:

REFLEXION IN THE X-AXIS:

[tex]h(x)=-f(x)[/tex]

REFLEXION IN THE Y-AXIS:

[tex]h(x)=f(-x)[/tex]

False by the same explanation as b.

The correct statement about the function y=sin(x) is that Reflection over either the x-axis or y-axis will change the graph. Therefore, option C is the correct answer.

The statement regarding the function y=sin(x) which is true is that reflection over either the x-axis or y-axis will change the graph.

This is because the sine function is an odd function, meaning that it has rotational symmetry about the origin. A characteristic of odd functions is that they satisfy the identity y(-x) = -y(x), not y(-x) = y(x), which describes an even function.

Therefore, the assumption Sin(x)=Sin(-x) would be incorrect, as it does not reflect the odd nature of the sine function. Thus, the correct answer is c. Reflection over either the x-axis or y-axis will change the graph.

In the attachement, there is what I came up with so far. I think that finding 'a' is non-trivial, if possible at all.

[tex]A_c[/tex] - the area of a circle

[tex]A_{cs}[/tex] - the area of a circular segment

Answer:

- the area of a circle

- the area of a circular segment

The simplified value is (x  - 6)

[tex]\frac{x^{2} -3x-18}{x+3}\\ = \frac{x^{2} -6x + 3x - 18}{x+3} \\=\frac{x(x-6) +3(x-6)}{x+3}\\ =\frac{(x+3)(x-6)}{x+3}\\ = x - 6[/tex]

So the simplified value is (x  - 6)

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To simplify the rational expression x^2 - 3x - 18 / x + 3, factor the numerator and cancel out the common factor (x + 3). The simplified form is x - 6.

Simplifying the Rational Expression

To simplify the expression
x² - 3x - 18 / x + 3, follow these steps:

(x - 6)(x + 3) / x + 3.

Next, cancel the common factor (x + 3):

(x - 6) (x + 3) / (x + 3) = x - 6

So, the simplified form of the expression is x - 6. Note that this simplification is valid for all values of x except -3, as the denominator would be zero.

for this case we have that by definition, the sum of the internal angles of a traingule is 180 degrees.

In addition, the angle "R" of the triangle is given by:

[tex]R = 180-45x[/tex]

So, we have to:

[tex](180-45x) + 25x + (57 + x) = 180\\180-45x + 25x + 57 + x = 180\\-45x + 25x + 57 + x = 0\\-45x + 25x + x = -57\\-19x = -57\\x = \frac {57} {19}\\x = 3[/tex]

Answer:

[tex]x = 3[/tex]

Answer with explanation:

It is given that for three integers , a, b and c, if

              [tex]\frac{ab}{c}\rightarrow then, \frac{a}{c} \text{or} \frac{b}{c}[/tex]

Since , a b is divisible by c , following are the possibilities

1.→ a and b are prime integers .Then , c will be prime number either equal to a or b.

2.→a and b are not prime integers ,then any of the factors of a or b will be equal to c.For example:

 ⇒a=m × n

 b=p × q× c

or,

⇒a=u×v×c

b=s×t

So, whatever the integral values taken by a, and b, if [tex]\frac{ab}{c}[/tex] then either of  [tex]\frac{a}{c} \text{or} \frac{b}{c}[/tex] is true.

Answer: 0.6767

Step-by-step explanation:

Given : Mean =[tex]\lambda=2[/tex] errors  per page

Let X be the number of errors in a particular page.

The formula to calculate the Poisson distribution is given by :_

[tex]P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}[/tex]

Now, the probability that a randomly selected page does not need to be retyped is given by :-

[tex]P(X\leq2)=P(0)+P(1)+P(2)\\\\=(\dfrac{e^{-2}2^0}{0!}+\dfrac{e^{-2}2^1}{1!}+\dfrac{e^{-2}2^2}{2!})\\\\=0.135335283237+0.270670566473+0.270670566473\\\\=0.676676416183\approx0.6767[/tex]

Hence, the required probability :- 0.6767

Answer:

  0.74

Step-by-step explanation:

P(A∪B) = P(A) + P(B) - P(A∩B) = 0.57 + 0.17 - 0

P(A∪B) = 0.74

The probability of A∩B is zero because the classes are mutually exclusive.

[tex]x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x[/tex]

Divide both sides by [tex]x^2[/tex]. In doing so, we force any possible solutions to exist on either [tex](-\infty,0)[/tex] or [tex]\boxed{(0,\infty)}[/tex] (the "positive" interval in such a situation is usually taken over the "negative" one) because [tex]x[/tex] cannot be 0 in order for us to do this.

[tex]\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x[/tex]

Condense the left side as the derivative of a product, then integrate both sides and solve for [tex]y[/tex]:

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac yx\right]=\sin x[/tex]

[tex]\dfrac yx=\displaystyle\int\sin x\,\mathrm dx[/tex]

[tex]\boxed{y=Cx-x\cos x}[/tex]

The general solution of a differential equation is to write y as a function of x.

Given

[tex]x \frac{dy}{dx} - y = x^2 \sin(x)[/tex]

Divide through by x

[tex]\frac{x}{x} \frac{dy}{dx} -\frac{y}{x} = \frac{x^2}{x} \sin(x)[/tex]

[tex]\frac{dy}{dx} -\frac{y}{x} = x \sin(x)[/tex]

Let P be function of x. Such that:

[tex]P(x) = -\frac 1x[/tex]

So, we have:

[tex]\frac{dy}{dx} +yP(x) = x\sin(x)[/tex]

Calculate the integrating factor I(x).

So, we have:

[tex]I(x) = e^{\int P(x) dx[/tex]

Substitute [tex]P(x) = -\frac 1x[/tex]

[tex]I(x) = e^{\int-\frac 1x dx[/tex]

Rewrite as:

[tex]I(x) = e^{-\int\frac 1x dx[/tex]

Integrate

[tex]I(x) = e^{-\ln(x)[/tex]

[tex]I(x) = \frac 1x[/tex]

So, we have:

[tex]\frac{dy}{dx} -\frac{y}{x} = x \sin(x)[/tex]

[tex][\frac{dy}{dx} -\frac{y}{x}] \frac 1x = [x \sin(x)] \frac 1x[/tex]

[tex][\frac{dy}{dx} -\frac{y}{x}] \frac 1x =\sin(x)[/tex]

Introduce [tex]I(x) = \frac 1x[/tex].

So, we have:

[tex]\frac{d}{dx}(\frac yx) = \sin(x)[/tex]

Multiply both sides by dx

[tex]d(\frac yx) = \sin(x)\ dx[/tex]

Integrate with respect to x

[tex]\frac yx = -\cos(x) + c[/tex]

Multiply through by x

[tex]y = -x\cos(x) + cx[/tex]

So, the general solution is: [tex]y = -x\cos(x) + cx[/tex], and the interval is [tex](0, \infty)[/tex]

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Answer:

  A)  -2

Step-by-step explanation:

The form is indeterminate at x=0, so L'Hopital's rule applies. The resulting form is also indeterminate at x=0, so a second application is required.

Let f(x) = x·sin(x); g(x) = cos(x) -1

Then f'(x) = sin(x) +x·cos(x), and g'(x) = -sin(x).

We still have f'(0)/g'(0) = 0/0 . . . . . indeterminate.

__

Differentiating numerator and denominator a second time gives ...

  f''(x) = 2cos(x) -sin(x)

  g''(x) = -cos(x)

Then f''(0)/g''(0) = 2/-1 = -2

_____

I like to start by graphing the expression to see if that is informative as to what the limit should be. The graph suggests the limit is -2, as we found.

Answer:

x < 33.84

Step-by-step explanation:

we have

13.48x-200 < 256.12

Solve for x

Adds 200 both sides

13.48x-200 +200 < 256.12+200

13.48x < 456.12

Divide by 13.48 both sides

13.48x/13.48 < 456.12/13.48

x < 33.84

The solution is the interval ----> (-∞, 33.84)

All real numbers less than 33.84

Answer:

The height of ball is 17 ft at t=0.55 and t=1.14.

Step-by-step explanation:

The general projectile motion is defined as

[tex]y=-16t^2+vt+y_0[/tex]

Where, v is initial velocity and y₀ is initial height.

It is given that the initial height is 7 and the initial upward velocity is 27.

Substitute v=27 and y₀=7 in the above equation to find the model for height of the ball.

[tex]h(t)=-16t^2+27t+7[/tex]

The height of ball is 17 ft. Put h(t)=17.

[tex]17=-16t^2+27t+7[/tex]

[tex]0=-16t^2+27t-10[/tex]

On solving this equation using graphing calculator we get

[tex]t=0.549,1.139[/tex]

[tex]t\approx 0.55,1.14[/tex]

Therefore the height of ball is 17 ft at t=0.55 and t=1.14.

[tex]|\Omega|=2^4=16\\|A|=4\\\\P(A)=\dfrac{4}{16}=\dfrac{1}{4}[/tex]

Answer:

0,008 or 0,8%

Step-by-step explanation:

To calculate the probability the selected password is made out only of lower-case letters, if it's only letters, we have first to find out how many passwords could be formed with only letters and with only lower-case letters.

For lowercase letters, we can make this many passwords, since for each of the 7  characters, we can pick among 26 lowercase letters:

NLL = 26 * 26 * 26 * 26 * 26 * 26 * 26

In the same fashion, for the number of passwords consisting only of letters, we can pick among 52 letters for each each character (26 lower-case, 26 upper-case):

NOL = 52 * 52 * 52 * 52 * 52 * 52 * 52

We can rewrite NOL differently to ease our calculations:

NOL = (2 * 26) * (2 * 26) * (2 * 26) * (2 * 26) * (2 * 26) * (2 * 26) * (2 * 26)

or

NOL = 26 * 26 * 26 * 26 * 26 * 26 * 26 * 2 * 2 * 2 * 2 * 2 * 2 * 2

Now we have to find out the probability a password containing only letters (NOL) is a password containing only lowercase letters (NLL).  So, we divide NLL by NOL:

[tex]\frac{NLL}{NOL} = \frac{26 * 26 * 26 * 26 * 26 * 26 * 26}{26 * 26 * 26 * 26 * 26 * 26 * 26 * 2 * 2 * 2 * 2 * 2 * 2 * 2}  = \frac{1}{2 * 2 * 2 * 2 * 2 * 2 * 2} = \frac{1}{2^{7} }[/tex]

The probability is thus 1/2^7 or 1/128 or 0,0078125

Which we are asked to round to 3 decimals... so 0,008 or 0,8%

Answer:

The solution is [tex]y(t)=e^{-t}(\cos 32t + (\frac{5}{8}) \sin 32t)[/tex]

Step-by-step explanation:

We need to find the solution of [tex]y''+2y'+17y=0[/tex] with

condition [tex]y(0)=3,\ y'(0)=17[/tex]

This is a homogeneous equation with characteristic polynomial

[tex]r^{2}+2r+17=0[/tex]

using quadratic formula [tex]x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/tex]

[tex]r=\frac{-2\pm \sqrt{2^{2}-4(1)(17)}}{2(1)}[/tex]

[tex]r=\frac{-2\pm \sqrt{4-68}}{2}[/tex]

[tex]r=\frac{-2\pm \sqrt{-64}}{2}[/tex]

[tex]r=\frac{-2\pm 64i}{2}[/tex]

[tex]r=-1 \pm 32i[/tex]

The general solution for eigen value [tex]a \pm ib[/tex] is

[tex]y(t)=e^{at}(A \cos bt + B \sin bt)[/tex]

[tex]y(t)=e^{-t}(A \cos 32t + B \sin 32t)[/tex]

Differentiate above with respect to 't'

[tex]y'(t)=-e^{-t}(A \cos 32t + B \sin 32t) + e^{-t}(-32A \sin 32t + 32B \cos 32t)[/tex]

Since, y(0)=3

[tex]y(0)=e^{0}(A \cos(0) + B \sin(0))[/tex]

[tex]3=(A \cos(0) +0)[/tex]

so, A=1

Since, y'(0)=17

[tex]y'(0)=-e^{0}(3 \cos(0) + B \sin(0)) + e^{0}(-32(3) \sin(0) + 32B \cos (0))[/tex]

[tex]17=-(3 \cos(0)) + (0 + 32B \cos (0))[/tex]

[tex]17=-3 + 32B[/tex]

add both the sides by 3,

[tex]17+3 = 32B[/tex]

[tex]20= 32B[/tex]

divide both the sides, by 32,

[tex]\frac{20}{32}= B[/tex]

[tex]\frac{5}{8}= B[/tex]

Put the value of constants in [tex]y(t)=e^{-t}(A \cos 32t + B \sin 32t)[/tex]

[tex]y(t)=e^{-t}((1) \cos 32t + (\frac{5}{8}) \sin 32t)[/tex]

Therefore, the solution is [tex]y(t)=e^{-t}(\cos 32t + (\frac{5}{8}) \sin 32t)[/tex]

y is (replace "is" with an equal sign) 4 less (replace with subtraction sign) than the product (multiply 5 and x) of 5 and x

y = 5x - 4

The reason the answer is like this ^^^ instead of y = 4 - 5x is because for this to be true it would have to say y is 5x less then 4

Hope this helped!

~Just a girl in love with Shawn Mendes

Answer:

y=5x-4

Step-by-step explanation:

Question about: ⇒ algebraic expression

Y: ⇒ Symbol into letters

is: ⇒ equal sign

less than: ⇒ <

product: ⇒ multiply

y=5x-4 is the correct answer.

I hope this helps you, and have a wonderful day!

Answer:

  C. zeros below the diagonal

Step-by-step explanation:

Upper echelon form (zeros below the diagonal) corresponds to a system of equations that has one equation in one variable, one equation in two variables, and additional equations in additional variables adding one variable at a time.

The single equation in a single variable is easily solved, and that result can be substituted into the equation with two variables (one of which is the one just found) to find one more variable's value. This back-substitution proceeds until all variable values have been found.

The process of producing such a matrix is called Gaussian Elimination.

__

The back-substitution process effectively makes the matrix be an identity matrix (diagonal = ones; zeros elsewhere) and the added column be the solution to the system of equations.

To solve a system of equations using the matrix method, you transform the augmented matrix to have zeros below the diagonal through Gaussian elimination. Then, you substitute back into the equations to find the solution.

To solve a system of equations using the matrix method, you use elementary row operations to transform the augmented matrix into one with zeros below the diagonal. This is achieved through a method called Gaussian elimination. The goal is to reduce the matrix to its row-echelon form, which leaves zeros below the diagonal. After this reduction, you can then proceed to substitute back into the equations to find the solution.

For example, let's take the system of equations:
x+2y=7
3x-4y=11
This can be represented as an augmented matrix:
[1 2 | 7]
[3 -4 | 11]
Using Gaussian elimination, we can eliminate the '3' below the diagonal by subtracting 3x the first row from the second, getting you:
[1 2 | 7]
[0 -10 | -10]
By substituting, we then find the solutions for the system of equations.

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Answer:

  102.5 g

Step-by-step explanation:

The mass is the sum of the values indicated by the sliders:

  100 g + 0 g + 2.5 g = 102.5 g

The mass, in grams, of the object being measured in the triple beam balance shown below is:

                               102.5 g

Triple Beam balance--

It is a instrument which is used to measure the mass of an object.

The advantage of using this device is that it measures the mass of an object precisely.

It has three counterweights in it.

One is of 100 gram, other is of 10 gram and the last is of 1 gram.

In order to find the mass of an object we add the weights in all the three sections.

The large slider is at 100 g, the medium slider is at 0 g, and the small slider is at 2.5 g.

                    100 g+0 g+2.5 g=102.5 g

Answer: 0.8413

Step-by-step explanation:

Given: Mean : [tex]\mu=\$40,500[/tex]

Standard deviation : [tex]\sigma = \$4,500[/tex]

The formula to calculate z-score is given by :_

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x= $36,000.00, we have

[tex]z=\dfrac{36000-40500}{4500}=-1[/tex]

The P-value = [tex]P(z\geq-1)=1-P(z<-1)=1-0.1586553=0.8413447\approx0.8413[/tex]

Hence, the probability that a randomly selected individual with an undergraduate business degree will get a starting salary of at least $36,000.00 = 0.8413

The probability of a randomly selected individual with an undergraduate business degree having a starting salary of at least $36,000, based on the given normal distribution with a mean of $40,500 and a standard deviation of $4,500, is approximately 0.8413 or 84.13%.

The question asks us to find the probability that a randomly selected individual with an undergraduate business degree will have a starting salary of at least $36,000.00, given that the mean starting salary is $40,500 with a standard deviation of $4,500. This problem can be solved using the properties of the normal distribution.

First, we calculate the z-score, which is the number of standard deviations away from the mean:

Z = (X - μ) / σ

Where X is the salary in question ($36,000), μ is the mean ($40,500), and σ is the standard deviation ($4,500). Plugging in the values:

Z = ($36,000 - $40,500) / $4,500 = -1

The next step is to look up this z-score in a standard normal distribution table or use a calculator with a standard normal distribution function to find the area to the right of this z-score. This area represents the probability we are looking for. Let's assume we found this area to be approximately 0.8413.

Therefore, the probability that a randomly selected individual with an undergraduate business degree will have a starting salary of at least $36,000 is about 0.8413 or 84.13%.

Answer:

88 ft²

Step-by-step explanation:

Area of larger square

10 × 8 = 80

10 × 8 because 10 is the length and 8 because the 6 and 2 rectangle is missing so it wouldn't be 10 × 10

4 × 2 = 8

4 × 2 = 8 because we need to work out the area of the smaller rectangle

80 + 8 = 88

Answer with explanation:

Given the differential equation

y''+y=0

The two function let

[tex]y_1= sint -cost[/tex]

[tex]y_2=sint+ cost[/tex]

Differentiate [tex]y_1 and y_2[/tex]

Then we get

[tex]y'_1= cost+sint[/tex]

[tex]y'_2=cost-sint[/tex]

Because [tex]\frac{\mathrm{d} sinx}{\mathrm{d} x} = cosx[/tex]

[tex]\frac{\mathrm{d}cosx }{\mathrm{d}x}= -sinx[/tex]

We find wronskin to prove that the function  is independent/ fundamental function.

w(x)=[tex]\begin{vmatrix} y_1&y_2\\y'_1&y'_2\end{vmatrix}[/tex]

[tex]w(x)=\begin{vmatrix}sint-cost&sint+cost\\cost+sint&cost-sint\end{vmatrix}[/tex]

[tex]w(x)=(sint-cost)(cost-sint)- (sint+cost)(cost+sint)[/tex]

[tex]w(x)=sintcost-sin^2t-cos^2t+sintcost-sintcost-sin^2t-cos^2t-sintcost[/tex]

[tex]w(x)=-sin^2t-cos^2t[/tex]    

[tex]sin^2t+cos^2t=1[/tex]

[tex]w(x)=-2\neq0[/tex]

Hence, the given two function are fundamental set of function on R.

Answer:

3t^2s

Step-by-step explanation:

15/3=5

t^5s/t^2s = t^3

t^2s^3/t^2s = s^2

For this case we have by definition, that a polynomial has a common factor when the same quantity, either number or letter, is found in all the terms of the polynomial.

We have the following expression:

[tex]3t ^ 5s-15t ^ 2s ^ 3[/tex]

So we have to:

[tex]3t ^ 2s[/tex] is the lowest common term in the terms of the expression:

[tex]3t ^ 2s (t ^ 3-5s ^ 2)[/tex]

Answer:

[tex]3t ^ 2s[/tex]

Answer:

100

Step-by-step explanation:

The value of the given logarithm is 100.

A logarithm is the power to which a number must be raised in order to get some other number.

Given that, log x = 2,

We will solve a logarithmic equation of x  by changing it to exponential form.

Now, the logarithmic equation is log₁₀x = 2

Since, we know that, logₐb = x then b = aˣ

Therefore, log₁₀x = 2

x = 10²

x = 100

Hence,  the value of the given logarithm is 100.

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Answer:

  30 connections

Step-by-step explanation:

20 switches with 3 connections each will have a total of 20×3 = 60 connections. That counts each connecting link twice, so only 30 connecting links are required.

Answer:

30 Connections!

Step-by-step explanation:

I did this on AoPs :)

Answer:

a) The expected value of M = 40

The standard error for M = 4

b) The expected value of M = 40

The standard error for M = 2

Step-by-step explanation:

* Lets revise some definition to solve the problem

- The mean of the distribution of sample means is called the expected

  value of M

- It is equal to the population mean μ

- The standard deviation of the distribution of sample means is called

  the standard error of M

- The rule of standard error is σM = σ/√n , where σ is the standard

  deviation and n is the size of the sample

* lets solve the problem

- A sample is selected from a population

∵ The mean of the population μ = 40

∵ The standard deviation σ = 8

a) The sample has n = 4 scores

∵ The expected value of M = μ

∵ μ = 40

∴ The expected value of M = 40

∵ The standard error of M = σ/√n

∵ σ = 8 and n = 4

∴ σM = 8/√4 = 8/2 = 4

∴ The standard error for M = 4

b) The sample has n = 16 scores

∵ The expected value of M = μ

∵ μ = 40

∴ The expected value of M = 40

∵ The standard error of M = σ/√n

∵ σ = 8 and n = 16

∴ σM = 8/√16 = 8/4 = 2

∴ The standard error for M = 2

When the sample has n = 4 scores then the expected value of M is 40 and the standard error of M is 4.

When the sample has n = 16 scores then the expected value of M is 40 and the standard error of M is 2.

Given

A sample is selected from a population with a mean of μ = 40 and a standard deviation of σ = 8. a. If the sample has n = 4 scores.

The mean of the distribution of sample means is called the expected value of M.

The standard deviation of the distribution of sample means is called the standard error of M.

1. The sample has n = 4 scores

The expected value of M = μ

The expected value of M = 40

The standard error of M is;

[tex]\rm Standard \ error=\dfrac{\sigma}{\sqrt{n} }\\\\ \sigma = 8 \ and \ n = 4}\\\\ Standard \ error=\dfrac{8}{\sqrt{4}}\\\\ Standard \ error=\dfrac{8}{2}\\\\ Standard \ error=4[/tex]

The standard error for M = 4

2.  1. The sample has n = 16 scores

The expected value of M = μ

The expected value of M = 40

The standard error of M is;

[tex]\rm Standard \ error=\dfrac{\sigma}{\sqrt{n} }\\\\ \sigma = 8 \ and \ n = 16}\\\\ Standard \ error=\dfrac{8}{\sqrt{16}}\\\\ Standard \ error=\dfrac{8}{4}\\\\ Standard \ error=2[/tex]

The standard error for M = 2

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Answer:

a) $40920

b) $38500

Step-by-step explanation:

Given:

5 employees,

Mean = $40300

Median = $38500

Min = $32000

If he lowest paid employee gets a $3100 raise, then his salary becomes

$32000+$3100=$35100

a) If the mean was $40300, then the sum of 5 salaries is

[tex]\$40300\cdot 5=\$201500[/tex]

After raising the lowest salary the sum becomes

[tex]\$201500+\$3100=\$204600[/tex]

and new mean is

[tex]\dfrac{\$204600}{5}=\$40920[/tex]

b) The  lowest salary becomes $35100. It is still smaller than the median, so the new median is the same as the old one.

New median = $38500

Answer: $64

Step-by-step explanation:

Set up is/of ratio. See photo attached. (:

The values of x for which the given vectors are basis for R³ is:

                        [tex]x\neq 1[/tex]

We know that for a set of vectors are linearly independent if the matrix formed by these set of vectors is non-singular i.e. the determinant of the matrix formed by these vectors is non-zero.

We are given three vectors as:

(-1,0,-1), (2,1,2), (1,1, x)

The matrix formed by these vectors is:

[tex]\left[\begin{array}{ccc}-1&2&1\\0&1&1\\-1&2&x\end{array}\right][/tex]

Now, the determinant of this matrix is:

[tex]\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-1(x-2)-2(1)+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+2-2+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+1[/tex]

Hence,

[tex]-x+1\neq 0\\\\\\i.e.\\\\\\x\neq 1[/tex]

ANSWER

The value of this function at x=-1 is 1

EXPLANATION

The given function is

[tex]f(x) = 3x + 4[/tex]

We want to find the value of this function at x=-1.

We substitute x=-1 into the function to obtain:

[tex]f( - 1) = 3( - 1)+ 4[/tex]

We multiply out to obtain:

[tex]f( - 1) = - 3+ 4[/tex]

[tex]f( - 1) = 1[/tex]

Therefore the value of this function at x=-1 is 1.

Answer:  [tex]f(-1)=1[/tex]

Step-by-step explanation:

Given the linear function f(x):

[tex]f(x)=3x+4[/tex]

By definition. a relation is a function if each input value has only one output value. In this case you need to find the output value for the input value [tex]x=-1[/tex]. In order to do this, you need to substitute this value of the variable "x" into the linear function given.

Then:

When [tex]x=-1[/tex]:

[tex]f(-1)=3(-1)+4[/tex]

Remember the multiplication of signs:

[tex](+)(-)=-\\(+)(+)=+\\(-)(-)=+[/tex]

Then, the value of f(x) when [tex]x=-1[/tex] is:

 [tex]f(-1)=-3+4[/tex]

 [tex]f(-1)=1[/tex]