During a manufacturing process, a metal part in a machine is exposed to varying temperature conditions. The manufacturer of the machine recommends that the temperature of the machine part remain below 132 ° F. The temperature T in degrees Fahrenheit x minutes after the machine is put into operation is modeled by T = -0.005x^2 + 0.45x + 125. Will the temperature of the part ever reach or exceed 132 ° F? Use the discriminant of a quadratic equation to decide.

A. no
B. yes

Answer:

B. Strech

Step-by-step explanation:

Stretching can change the shape of something such as a function.

Answer:

C.

Step-by-step explanation:

We are going to solve it with substitution, that is, we are going to clear one variable and then replace it in the other equation. Then,

[tex]3x + 2y = 12[/tex]

[tex]3x = 12-2y[/tex]

[tex]x = \frac{12-2y}{3}[/tex]

[tex]x = \frac{12}{3}-\frac{2y}{3}[/tex]

[tex]x = 4-\frac{2y}{3}.[/tex]

Now, we replace that x value in the equation 6x + 3y = 21:

[tex]6(4-\frac{2y}{3}) + 3y = 21[/tex]

[tex]24-\frac{12y}{3} + 3y = 21[/tex]

[tex]24-4y + 3y = 21[/tex]

[tex]24-y = 21[/tex]

[tex]-y = 21-24[/tex]

[tex]-y = -3[/tex]

[tex]y = 3.[/tex]

Finally, we replace the y value found to find x.

[tex]x = 4-\frac{2*3}{3}[/tex]

[tex]x = 4-2[/tex]

[tex]x = 2.[/tex]

So, the solution is (x,y)=(2,3). Then, the answer is C.

Answer: The correct option is A.

Explanation:

The given function is,

[tex]f(x)=2(0.4)^x[/tex]

To find the graph of this function after the reflecting across y-axis, first we have to find the graph of the equation.

The value of the function is 2 when x=0, so, the graph of given equation intersect the y-axis at 2.

In the equation [tex](0.4)^x[/tex]. Since [tex]0<0.4<1[/tex], so the given function is decreasing function.

[tex]f(x)\rightarrow 0 \text{ as }\rightarrow \infty[/tex]

[tex]f(x)\rightarrow \infty \text{ as }\rightarrow -\infty[/tex]

The value of f(x) is always positive, so the graph of f(x) is always above the x-axis. Thus, the graph must be above the x-axis after reflection across y-axis.

So, the option (2) and (4) and incorrect.

When we reflect the graph across the y-axis then,

[tex]f(x)\rightarrow \infty \text{ as }\rightarrow \infty[/tex]

[tex]f(x)\rightarrow 0 \text{ as }\rightarrow -\infty[/tex]

It means when x approaches to large negative number the f(x) approaches to 0 and when x approaches to large positive number the f(x) approaches to infinite.

Therefore, the correct option is show in first graph.

Answer:

a

Step-by-step explanation:

answer: (f-i)/-g=t

I hope it helps~

The correct option is A: [tex]\( \boxed{f(x) = 3(8)^x} \).[/tex]

To determine which option is equivalent to the given function f(x) = [tex]3(2)^{3x} \),[/tex] we will simplify the given function and compare it to each option.

The given function is:

[tex]\[ f(x) = 3(2)^{3x} \][/tex]

We can rewrite the exponent part using properties of exponents:

[tex]\[ (2)^{3x} = (2^3)^x = 8^x \][/tex]

So the function becomes:

[tex]\[ f(x) = 3 \cdot 8^x \][/tex]

Now, let's compare this to each of the given options:

Option A: [tex]\( f(x) = 3(8)^x \)[/tex]

This is exactly the same as our simplified function. Therefore, Option A is equivalent to [tex]\( f(x) = 3(2)^{3x} \).[/tex]

Option B: [tex]\( f(x) = 24^x \)[/tex]

This is not equivalent because it implies a different base raised to the power of x.

Option C:[tex]\( f(x) = 3(8x) \)[/tex]

This is not equivalent because it multiplies by 27 instead of 3.

Option D: [tex]\( f(x) = 3(8x) \)[/tex]

This is not equivalent because it implies multiplying 8 by x, not raising 8 to the power of x.

Therefore, the correct answer is:

[tex]\( \boxed{f(x) = 3(8)^x} \).[/tex]

The complete question is here:

Which of the following options is an equivalent function to f(x) = 3(2)3x?

A. f(x) = 3(8)x

B. f(x) = 24x

C. f(x) = 27(8)x

D. f(x) = 3(8x)

Answer:

The equation has one real root and two complex roots.

Step-by-step explanation:

The given equation is

[tex]x^3-3x^2+4x-12=0[/tex]

The above equation is true form x=3, therefore (x-3) is a factor of above equation.

Use long division or synthetic division method to divide the equation by (x-3).

[tex](x-3)(x^2+4)=0[/tex]

Equate each factor equal to zero.

[tex]x-3=0[/tex]

[tex]x=3[/tex]

Therefore 3 is a real root of the equation.

[tex]x^2+4=0[/tex]

[tex]x^2=-4[/tex]

[tex]x=\sqrt{-4}[/tex]

[tex]x=\pm 2i[/tex]

2i and -2i are complex roots of the equation.

Therefore the equation has one real root and two complex roots.

A factory producing 1200 pounds of chips in four hours will yield 3000 pounds when operating for 10 hours, assuming a consistent production rate.

The given problem pertains to rate, which allows the determination of how much of something can be done in a specific amount of time. In this case, the rate of potato chip production is 1200 pounds every four hours. To find out how many pounds of potato chips can be produced in 10 hours, we first need to determine the hourly rate of production.

Firstly, we determine the hourly production rate of the factory by dividing the total produce by the time spent. That is 1200 pounds / 4 hours = 300 pounds/hour.

At this hourly rate, the amount of potato chips that can be produced in 10 hours can be calculated by multiplying the hourly rate by the number of hours, which is 300 pounds/hour * 10 hours = 3000 pounds.

So, a factory that produces 1200 pounds of potato chips in four hours will make 3000 pounds of potato chips in 10 hours.

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The valid postulates and theorems for proving triangles congruent are ASA, AAS, and SAS. The SSA postulate is not a valid postulate.

The valid postulates and theorems for proving triangles congruent are ASA (Angle-Side-Angle), AAS (Angle-Angle-Side), and SAS (Side-Angle-Side).

The SSA (Side-Side-Angle) postulate, also known as the ambiguous case of the Law of Sines, is not a valid postulate for proving triangles congruent because it can result in two different triangles or no triangle at all.

Thus, the correct answer is A. SSA.

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Answer:

1. Which equation can be used to find the time to complete the job when x workers are assigned to it?

A.) m(x) = 270/x

2. Suppose that H(x) varies inversely with x and H(x)=50 when x =0.25

What is H(x) when x =2?

B.) 6.25

Step-by-step explanation:

1. We know when x=3 (workers), m(x)= 90 min (time is takes to mow the lawn at a large park) and we know m(x) varies inversely with x, this means when m(x) increases so x decreases. This is obtained with a division over x.

Now, we can replace to know what is the correct answer:

[tex]m(x)=\frac{y}{x} (1) \\90=\frac{y}{3} \\90*3=y[/tex]

[tex]y=270[/tex]

The answer is A.

[tex]m(x)=\frac{270}{x}[/tex]

We can confirm if we replace x=3 in the equation before. Our answer should be 90.

[tex]m(x)=\frac{270}{3}=90[/tex]

2.  Suppose that H(x) varies inversely with x and H(x)=50 when x =0.25

We know H(x) varies inversely with x, this means when H(x) increases so x decreases. This is obtained with a division over x. Now, we need to find the constant in the numerator.

Now, we can replace in the eq (1) to know the constant in the numerator:

[tex]H(x)=\frac{y}{x}\\50=\frac{y}{0.25} \\50*0.25=y[/tex]

[tex]y=12.5[/tex]

Now, we can replace in the eq (1) to find H(x) when x=2

[tex]m(x)=\frac{y}{x}[/tex]

[tex]m(x)=\frac{12.5}{2}[/tex]

[tex]m(x)=6.25[/tex]  

The answer is B.) 6.25

Answer:

It’s actually 5

Step-by-step explanation:

Ape x

Using the point slope equation, it is determined that Keisha will not finish reading the 325 pages book in 10 days if she is reading at a rate of 25 pages per day.

The subject of this question is Mathematics and it involves solving a problem using a point slope equation.

Let's denote the total number of pages in the book as 'y' and the number of pages that Keisha can read per day as 'm'. We know that y=325 and m=25. The point slope equation in this case would be y=mx, where 'x' represents the number of days. In this case, we are trying to find out whether x=10.

Substituting the known values into the equation we get: 325=25*10. This simplifies to 325=250.

Since 325 does not equal 250, we can conclude that Keisha will not complete reading the book in 10 days.

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It take 6.434 month  or  6 month and 13 days you to accumulate a balance of $1000 in your account.

Word problem is a mathematical problem totally described in language that is frequently used as an educational tool.

Your current account balance is $215.

Your per month income is $444.

Your per month expense is $322.

Hence, your per month savings is = $444 - $322 = $122.

Let, it take x month you to accumulate a balance of $1000 in your account.

Then mathematically,

215 + 122 x = 1000

Rearrange the equation can be written that:

122 x = 1000 -215

After the subtraction, the result will be:

122 x = 785

Rearrange the equation can be written that:

x = 785/122

After the division, the result will be:

x = 6.434 = 6 month and 13 days.

Hence,  it take 6.434 month  or  6 month and 13 days you to accumulate a balance of $1000 in your account.

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Answer:

It would take 2 seconds to the key to reach the ground.

The domain the function is: 0≤t≤2 or in interval notation [0,2].

The range of the function is: 0≤h≤64 or in interval notation [0,64].

Step-by-step explanation:

The provided function is: [tex]h=-16t^2 +64[/tex]

Where h represents the height and t represents the time.

Part(A) how long does it take the key to reach the ground?

When key will be at ground, the height will be 0.

Substitute the value of h=0 in above equation and solve for t.

[tex]-16t^2 +64=0[/tex]

[tex]-16(t^2 -4)=0[/tex]

[tex](t^2 -2^2)=0[/tex]

[tex](t+2)(t-2)=0[/tex]

[tex]t+2=0\ or\ t-2=0[/tex]

[tex]t=-2\ or\ t=2[/tex]

Ignore the negative value of t.

Hence, it would take 2 seconds to the key to reach the ground.

Part (B) what are the reasonable domain and range for the function h?

Time and Distance should be a positive number.

Thus, the value of t and h should be greater or equal to 0.

From the part (a) we know that the after 2 second the key will hit the ground so the value of t can be greater or equal to 0 but should be less or equal to 2.

Thus, the domain the function is: 0≤t≤2 or in interval notation [0,2].

The maximum height of the key will be at t=0.

Thus the maximum value of h can be 64 and minimum value can be 0.

The range of the function is: 0≤h≤64 or in interval notation [0,64].