1/2 = 3/6, shade the model to find

Mathematics

2 answers:

Elena L [17]3 years ago

8 0

You would just Shade half of the model.

Monica [59]3 years ago

3 0

Just shade half of the problem if it says to shade 1/2 then shade the half doesn't matter what side

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What is the top answers thank you

bulgar [2K]

2 because the X and 2 are being multipled at the top so im assuming thats what its going by

4 0

3 years ago

Which of the following options results in a graph that shows exponential growth?

poizon [28]

Answer:

$f(x)=0.7(5)^x$

Step-by-step explanation:

$\text{Exponential function: }y=a\cdot b^x$

It is two types of function.

Exponential Growth: If function increase as value of x increase.

In this case, a>0 and b>1

Exponential decay: If function decrease as value of x increase.

In this case, a>0 and 0<b<1

We need to choose exponential growth function.

$\text{Exponential Growth function: }f(x)=0.7(5)^x$

Here a=0.75>0 and b=5>1

Therefore, This function is exponential growth.

7 0

3 years ago

Read 2 more answers

How do you integrate arctan(x dx? i think that if you simplify the integral you get:?

Temka [501]

Integration by parts will help here. Letting $u=\arctan x$ and $\mathrm dv=\mathrm dx$ , you end up with $\mathrm du=\dfrac{\mathrm dx}{1+x^2}$ and $v=x$ . Now

$\displaystyle\int\arctan x\,\mathrm dx=uv-\int v\,\mathrm du$
$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx$

For the remaining integral, setting $y=1+x^2$ gives $\dfrac{\mathrm dy}2=x\,\mathrm dx$ , so

$\displaystyle\int\frac x{1+x^2}\,\mathrm dx=\frac12\int\frac{\mathrm dy}y=\frac12\ln|y|+C=\frac12\ln(1+x^2)+C$

Putting everything together, you end up with

$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C$