What is the solution to the equation below? 2(x - 3) = 2x 5?

The probability that exactly six children in the family are boys is 0.015625.

An arrangement of objects where the order in which the objects are selected does not matter.

The probability of having exactly six boys in a family with eight children can be calculated using the binomial distribution formula:

[tex]P(X = k) =^nC_{k} p^k(1-p)^(^n^-^k^)[/tex]

P(X = k) is the probability of having exactly k boys, n is the total number of children,  k is the number of boys we want to find , p is the probability of having a boy and  (1-p) is the probability of having a girl (also 1/2 in this case)

Substituting the values into the formula, we get:

P(X = 6) = ⁸C₆ (1/2)⁶ (1/2)⁸⁻⁶

= (8! / (6! ×2!)) (1/2)⁸

= (87/21) × (1/2)^8

= 0.015625

Therefore, the probability that exactly six children in the family are boys is 0.015625.

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Brandi rides her bike at a speed of 5 m/s from school to her house. She will take an additional 188 seconds to cover the remaining distance of 940 meters to arrive at her house.

Brandi's situation:

To find: Additional time needed to reach house.

Calculations:

Answer:

D. LG = MN

Step-by-step explanation:

Final answer:

To prove two triangles congruent by SAS, we require equality of two sides and the included angle. Given line LT = line MQ and angle L = angle M, we also need line GT = line NQ and line LG = line MN. For the second part, line RS = line WU and line RT = line VU are necessary to fulfill SAS criteria.(Options C and D for first and Option E and F for second)

Explanation:

When proving two triangles are congruent using the Side-Angle-Side (SAS) postulate, you need to know that two sides and the included angle (the angle between the two sides) of one triangle are exactly equal to two sides and the included angle of another triangle.

In the given problem, we're told that line LT is congruent to line MQ and angle L is congruent to angle M. The additional information needed to prove triangle congruence by SAS would be to show that the second pair of sides around the included angle are also congruent (as in option C Line GT = line NQ) and to ensure congruence on the corresponding parts of the other triangle (as in option D Line LG = line MN).

Answering the second part, to prove congruence by SAS, we would need two corresponding sides and the included angle. Therefore, the correct choices are option E Line RS = line WU which provides the second side, and option F line RT = line VU ensures the sides are corresponding in the two triangles, enclosing the angle which is already given as equal.

The situation represents the linear function.

Given that,

Based on the above information, we can say that

When dollar presents y value and the quarter presents x value

So, the slope should be [tex]\frac{1}{4}[/tex]

The linear function should be [tex]y = \frac{1}{4}x[/tex]

Therefore we can conclude that the situation represents the linear function.

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The area of the real rectangular dining hall (assuming that 5cm on paper is 6m in real) is 72m².

A rectangle is a quadrilateral whose opposite edges are of equal length and the opposite angles are of equal measurements.

Given that

the length of the dining hall, l = 10cm

the width of the dining hall, b = 5cm

As evident from the figure, the dining hall is rectangular, and since it is known that the area of a rectangle is the product of its length and width, therefore,

Area of the hall (on paper) = l × b

                          = 10 × 5

                          = 50 square cm.

Also, as it is given in the question that 5cm on paper represents 6m in real, therefore

[tex]Area\ of\ the\ hall\ (in\ real)= (\dfrac{10}{5}\times6)\times(\dfrac{5}{5}\times6)[/tex]

                                          [tex]\\= 2\times6\times6\\= 72\ m^2[/tex]

Hence, the area of the hall in real is 72m².

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The given sequence is in the form of a Riemann sum for the integral of the function 1/(1+x) from 1 to 2. So, the correct answer is b) 'integral from 1 to 2 of (1/(x+1)) dx'.

The sequence in the question seems to be in the form of a Riemann sum for an integral. The term inside the summation loop can be expressed as 1/(1+i/n) where i varies from 1 to n. The limit of the sequence as n tends to infinity can be thus represented as the integral from 0 to 1 of the function 1/(1+x) dx. The function 1/(1+x) is a continuous function over the interval [0,1], so the integral exists and the limit of the sequence is legitimately defined.

Therefore, looking in the options provided, the answer to your question will be 'b.) integral from 1 to 2 of (1/(x+1)) dx'. This integral is the limit of the given series as n approaches infinity.

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The limit as 'n' approaches infinity of this summation is equal to the integral from 1 to 2 of (1/(x+1))dx. Essentially, this question is turning the sum into an integral as 'n' approaches infinity.

This question is asking for the limit of a sum as n approaches infinity, which is essentially the definition of an integral. Every term in the sum could be expressed as 1/(1+i/n), where 'i' is the sum index. Each term, when 'n' becomes infinitely large, becomes a term of the form i/n, or Δx, which is a small piece of the variable we’re integrating. Similarly, each 1/(1+i/n) term turns into 1/x for that small piece.

Given these transformations and the fact that the index i ranges from 1 to n, it's clear that as n approaches infinity, we are integrating the function 1/x from 1 to 2.

Thus, the correct answer would be: b.) integral from 1 to 2 of (1/(x+1))dx

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First line because 4(2a+1)=8a+4 and 9a+4-(8a+4)= a

Second line because a times 2 =2a and 2a+1-2a=1

Although the second equality is more or less obvious since 2a+1 leaves a remainder of 1 when divided by a.

Answer with explanation:

The given fourth degree polynomial is:

  [tex]f(x)=x^4+6 x^3-3 x^2+17 x-15[/tex]

By rational root theorem , the possible Zeroes of the polynomial are factors of 15 .

Factors of 15 are

  [tex]=\pm 1, \pm 3, \pm 5, \pm 15[/tex]

There are 8 possible Zeroes of the given Polynomial expression.

Answer:

Amount of C-14 taken were 2.5 pounds and 3.5 pounds respectively.

Step-by-step explanation:

Radioactive decay is an exponential process represented by

[tex]A_{t}=A_{0}e^{-kt}[/tex]

where [tex]A_{t}[/tex] = Amount of the radioactive element after t years

[tex]A_{0}[/tex] = Initial amount

k = Decay constant

t = time in years

Half life period of Carbon-14 is 5730 years.

[tex]\frac{A_{0} }{2}=A_{0}e^{-5730k}[/tex]

[tex]\frac{1}{2}=e^{-5730k}[/tex]

Now we take ln (Natural log) on both the sides

[tex]ln(\frac{1}{2})=ln[e^{-5730k}][/tex]

-ln(2) = -5730kln(e)

0.69315 = 5730k

[tex]k=\frac{0.69315}{5730}[/tex]

[tex]k=1.21\times 10^{-4}[/tex]

Now we have to calculate the weight of samples of C-14 taken for the remaining quantities [tex]\frac{5}{8}[/tex] and [tex]\frac{7}{8}[/tex] of a pound.

[tex]\frac{5}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}[/tex]

[tex]\frac{5}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}[/tex]

[tex]\frac{5}{8}=A_{0}e^{(-1.3863)}[/tex]

[tex]A_{0}=\frac{5}{8}\times e^{1.3863}[/tex]

[tex]A_{0}=\frac{5}{8}\times 4[/tex]

[tex]A_{0}=\frac{5}{2}[/tex]

[tex]A_{0}=2.5[/tex] pounds

Similarly for [tex]\frac{7}{8}[/tex] pounds

[tex]\frac{7}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}[/tex]

[tex]\frac{7}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}[/tex]

[tex]\frac{7}{8}=A_{0}e^{(-1.3863)}[/tex]

[tex]A_{0}=\frac{7}{8}\times e^{(1.3863)}[/tex]

[tex]A_{0}=\frac{7}{8}\times 4[/tex]

[tex]A_{0}=\frac{7}{2}[/tex]

[tex]A_{0}=3.5[/tex] pounds

The equation 5x^2 + 5y^2 + 10x - y = 0 represents a circle. By completing the square, the equation is rewritten in the standard form as (x + 1)^2 + (y - 1/10)^2 = 101/100. The circle's center is at (-1, 1/10) with a radius of approximately 1.005.

To show that the equation 5x^2 + 5y^2 + 10x - y = 0 represents a circle and to rewrite it in standard form, we can complete the square for both x and y terms. First, factor out the coefficients of the squared terms:

Divide both sides by 5 to simplify the equation:

Now, complete the square by adding and subtracting the necessary constants inside each parenthesis:

By completing the square, the equation becomes:

This can be further simplified to:

The standard form of the equation for a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius of the circle. Therefore, our circle's center is at (-1, 1/10) and its radius is the square root of 101/100, which simplifies to 10.05/10 or approximately 1.005.

Answer:

Step-by-step explanation:

Simple interest formula is : [tex]p\times r\times t[/tex]

t is always in years here.

1.

p = 250

r = 2.7% or 0.027

t = 1

Simple interest earned = [tex]250\times0.027\times1[/tex] = $6.75

2.

p = 389.42

r = 3.2% or 0.032

t = 7

Simple interest earned = [tex]389.42\times0.032\times7[/tex] = $87.23

Amount after 7 years will be = [tex]389.42+87.23[/tex] = $476.65

3.

p = 1567.12

r = 1.9% or 0.019

t = [tex]9/12=0.75[/tex]

Simple interest earned = [tex]1567.12\times0.019\times0.75[/tex] = $22.33

Account balance after 9 months = [tex]1567.12+22.33[/tex] = $1589.45

Answer:

A.

Step-by-step explanation:

The value of x is 7/6.

To find the value of x, we can utilize the given information about the lengths of the medians and segments within triangle ABC.

Using medians CE and GC:

Since G is the centroid of triangle ABC, it divides each median into two segments in a ratio of 2:1. Therefore, we can set up two equations based on the given lengths:

CE = 2/3 * GC

6x = 2/3 * (3x + 7)

Solving for x, we get:

6x = 2x + 14/3

4x = 14/3

x = 7/6

Using medians BG and FG:

Similarly, we can set up two equations based on the given lengths of medians BG and FG:

FG = 2/3 * BG

x + 8 = 2/3 * (5x - 1)

Solving for x, we get:

x + 8 = 10x/3 - 2/3

8/3 = 9x/3

x = 8/9

Using medians DG and AG:

Following the same approach, we can set up two equations based on the given lengths of medians DG and AG:

AG = 2/3 * DG

9x - 6 = 2/3 * (4x - 5)

Solving for x, we get:

9x - 6 = 8x/3 - 10/3

x - 6 = 8x/3 - 10/3

-5 = 5x/3

x = -3

Comparing the values obtained from each set of equations, we find that x = 7/6 is consistent across all three sets. Therefore, the value of x is 7/6.

The fifth term in the binomial expansion of [tex](x+5)^{8}[/tex] is [tex]\boxed{43,750\ x^{4}}[/tex].

Further explanation:

Given:

The binomial term is [tex](x+5)^{8}[/tex].

The expansion of [tex](x+5)^{8}[/tex] is as follows:

[tex]\boxed{{\left({a+b}\right)^n}=\sum\limits_{k=0}^n{{}^n{{\text{C}}_k}{a^{n - k}}{b^k}}}[/tex]

There are [tex]n+1[/tex] terms in the expansion of [tex](a+b)^{n}[/tex].

The sum of indices of [tex]a[/tex] and [tex]b[/tex] is equal to [tex]n[/tex] in every term of the expansion.

The general term [tex]T_{r+1}[/tex] of the binomial term [tex](a+b)^{n}[/tex] is as follows:

[tex]\boxed{{{\text{T}}_{r + 1}}={}^n{{\text{C}}_r}{a^{n - r}}{b^r}}[/tex]

For [tex]5^{th}[/tex] term the value of [tex]r[/tex] is calculated as follows:

[tex]\begin{aligned}r+1&=5\\r&=5-1\\r&=4\end{aligned}[/tex]

Now, the [tex]5^{th}[/tex] term of [tex](x+5)^{8}[/tex] is calculated as follows:

[tex]\begin{aligned}T_{5}&=T_{4+1}\\&=^8C_{4}\cdot x^{8-4}\cdot 5^{4}\\&=\dfrac{8\cdot 7\cdot 6\cdot 5}{4\cdot 3\cdot 2\cdot 1}\cdot x^{4}\cdot 625\\&=625\cdot 70x^{4}\\&=43,750x^{4}\end{aligned}[/tex]

Therefore, the fifth term of the binomial expansion [tex](x+5)^{8}[/tex] is [tex]\boxed{43,750\ x^{4}}[/tex].

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Answer details:

Grade: Senior school

Subject: Mathematics

Chapter: Binomial Theorem

Keywords: Binomial theorem, expansion, (x+5)^8, 175000x3, 43750x4, 3125x5, 7000x5, fifth term, binomial expansion, genral term, binomial, polynomial, indices.

The answer is:

[tex]\[\boxed{43750x^4}\][/tex]

To find the fifth term in the binomial expansion of [tex]\((x + 5)^8\)[/tex], we use the binomial theorem. The binomial theorem states that:

[tex]\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\][/tex]

For the expansion of [tex]\((x + 5)^8\)[/tex], we identify:

[tex]\[a = x, \quad b = 5, \quad n = 8\][/tex]

The general term in the expansion is given by:

[tex]\[T_{k+1} = \binom{n}{k} a^{n-k} b^k\][/tex]

We need to find the fifth term, which corresponds to  k = 4  (since k  starts from 0):

[tex]\[T_5 = \binom{8}{4} x^{8-4} 5^4\][/tex]

First, calculate the binomial coefficient [tex]\(\binom{8}{4}\):[/tex]

[tex]\[\binom{8}{4} = \frac{8!}{4! \cdot 4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70\][/tex]

Next, calculate [tex]\( x^{8-4} \)[/tex] and [tex]\( 5^4 \):[/tex]

[tex]\[x^{8-4} = x^4\]\[5^4 = 625\][/tex]

Now, combine these results:

[tex]\[T_5 = 70 \cdot x^4 \cdot 625\][/tex]

Finally, multiply the coefficients:

[tex]\[70 \times 625 = 43750\][/tex]

Thus, the fifth term is:

[tex]\[43750x^4\][/tex]

Therefore, the answer is:

[tex]\[\boxed{43750x^4}\][/tex]

Answer:

(<2 and <4 are vert angles)     reason:given

(lines m and n intersect at p)   reason: def of vertical angles

(<2 and <3 are a linear pair)      reason: def of a linear pair

(<2 and <3 are a linear pair)      reason: def of a linear pair

(m<2+m<3=180)                          reason:angle addition postulate

(m<3+m<4=180)                          reason:angle addition postulate

(m<2+m<3=m<3+m<4                  reason:substitution property

(m<2 = m<4)                                 reason:subtraction property

<2 ~<4                                            reason:definition of ~ angles

Step-by-step explanation:

edgenuity 2020

The function y=x-cosx has an extremum on the interval [0,2pi] at x=3pi/2, determined by setting the first derivative equal to zero. The second derivative test is inconclusive at this point, but the point (3pi/2, 3pi/2) is a minimum due to the nature of the function.

To find all points of extrema on the interval [0,2pi] for the function y=x-cosx, we first need to find the derivative of the function and determine where it is equal to zero. The first derivative of y=x-cosx is y' = 1 + sinx. The critical points occur where y' = 0, which translates to 1 + sinx = 0 or sinx = -1. The only point in the interval [0,2pi] where sinx = -1 is at x = 3pi/2. To determine if this critical point is a maximum or minimum, we apply the second derivative test.

The second derivative of the function is y'' = cosx. At x = 3pi/2, the second derivative y'' = cos(3pi/2) = 0; since the second derivative is zero, the second derivative test is inconclusive. However, we can analyze the function around the critical point to conclude or use other methods like analyzing the first derivative's sign change.

For y=x-cosx, since the original function is a combination of a linear increasing function and cosine function, the critical point at x = 3pi/2 will be a minimum as the function increases both before and after this point.

Therefore, the point of extrema on the interval [0,2pi] for the function y=x-cosx are at (3pi/2, 3pi/2 - cos(3pi/2)), which simplifies to (3pi/2, 3pi/2).