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3 years ago

Estimate the average rate of change between x = 0 and x = 2 for the function shown.

Mathematics

1 answer:

Dovator [93]3 years ago

7 0

Answer:

the answer to this question is b

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A rocket is launched into the air and follows the path, h (t) = -3 * t^2 + 12 * t

DanielleElmas [232]

check the picture below.

so, the rocket will come back to the ground when h(t) = 0, thus

$\bf h(t)=-3t^2+12t\implies \stackrel{h(t)}{0}=-3t^2+12t\implies 0=-3t(t-4)\\\\[-0.35em]~\dotfill\\\\0=-3t\implies 0=t\impliedby \textit{0 seconds when it took off from the ground}\\\\[-0.35em]~\dotfill\\\\0=t-4\implies 4=t\impliedby \textit{4 seconds later, it came back down}$

8 0

4 years ago

Simplify: (picture)<br> and please show how you got to the answer, thanks!

Troyanec [42]

$(16 a^{3} b^{4} )^ \frac{3}{4}$

Expand,

$16^{ \frac{3}{4} } a^{3* \frac{3}{4} } b^{4* \frac{3}{4 } }$
= $8 a^{ \frac{9}{4} } b^{3}$

So the third choice would be your answer!

Hope this helps!

6 0

3 years ago

In a school,2/3 of the students study a language.

Cloud [144]

Answer:

The ratio is 4/11

Step-by-step explanation:

Let

x ----> the total number of students in a school

we know that

1) 2/3 of the students study a language.

(2/3)x ----> number of students studying a language

and

x-(2/3)x=(1/3)x ---->number of students who do not study a language

2) Of those who study a language 2/5 were studying french

(2/3)x(2/5)=(4/15)x ----> number of students studying French

and

(2/3)x(3/5)=(6/15)x ----> number of students studying a language but not studying French

3) we know that

The number of students who do not study French is equal to the number of students who do not study a language plus the number of students who study a language but do not study French.

(1/3)x+(6/15)x=(11/15)x ----> number of students who not study French

(4/15)x ----> number of students who study French

4 ) Find the ratio of students who study french and who do not

(4/15)x/(11/15)x=4/11

7 0

3 years ago

Integrate the following problem:

vazorg [7]

Answer:

$\displaystyle \frac{2 \cdot sin2x-cos2x}{5e^x} + C$

Step-by-step explanation:

The integration by parts formula is: $\displaystyle \int udv = uv - \int vdu$

Let's find u, du, dv, and v for $\displaystyle \int e^-^x \cdot cos2x \ dx$ .

$u=e^-^x$
$du=-e^-^x dx$

$dv=cos2x \ dx$
$v= \frac{sin2x}{2}$

Plug these values into the IBP formula:

$\displaystyle \int e^-^x \cdot cos2x \ dx = e^-^x \cdot \frac{sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx$
$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx$

Now let's evaluate the integral $\displaystyle \int \frac{sin2x}{2} \cdot -e^-^x dx$ .

Let's find u, du, dv, and v for this integral:

$u=-e^-^x$
$du=e^-^x dx$
$dv=\frac{sin2x}{2} dx$
$v=\frac{-cos2x}{4}$

Plug these values into the IBP formula:

$\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} - \int \frac{-cos2x}{4}\cdot e^-^x dx$

Factor 1/4 out of the integral and we are left with the exact same integral from the question.

$\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx$

Let's substitute this back into the first IBP equation.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]$

Simplify inside the brackets.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ \frac{e^-^x \cdot cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]$

Distribute the negative sign into the parentheses.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4} - \frac{1}{4} \int cos2x \cdot e^-^x dx$

Add the like term to the left side.

$\displaystyle \int e^-^x \cdot cos2x \ dx + \frac{1}{4} \int cos2x \cdot e^-^x dx= \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}$
$\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}$

Make the fractions have common denominators.

$\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x}{4} - \frac{e^-^x \cdot cos2x}{4}$

Simplify this equation.

$\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4}$

Multiply the right side by the reciprocal of 5/4.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4} \cdot \frac{4}{5}$

The 4's cancel out and we are left with:

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{5}$

Factor $e^-^x$ out of the numerator.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x(2 \cdot sin2x-cos2x)}{5}$

Simplify this by using exponential properties.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2 \cdot sin2x-cos2x}{5e^x}$

The final answer is $\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2 \cdot sin2x-cos2x}{5e^x} + C$ .

7 0

3 years ago

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What is the length of the line segment BC

Blizzard [7]

I don't know you have to use a measuring tool to figure it out ruler helps sorry try a ruler of a angle scale

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3 years ago

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