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3 years ago

Find the area, round to nearest tenth

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

3 0

Answer:

I think it would be 350

Step-by-step explanation:

25x7=175

175/2=87.5

87.5 would be the area of 1 triangle. There are 4 triangles so

87.5x4=350

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Simplify: 3 square root 216 A. 8 B. 16 C. 6 D. 2

zysi [14]

Answer:

Step-by-step explanation:

Use a calculator

4 0

3 years ago

Read 2 more answers

Can someone help me find the volume and the surface area of this shape.

mash [69]

Answer:

Got my points deducted by a loser name PoeticAesthetics

Step-by-step explanation:

5 0

2 years ago

J/2--12=15 j/2 is a fraction

Natasha_Volkova [10]

Answer:

j = 54

Step-by-step explanation:

j/2 - 12 = 15

j/2 = 15 +12

j/2 *2 = 27 * 2

j = 54

If you're kind enough, please mark this as the brainlist answer!

3 0

3 years ago

"find the reduction formula for the integral" sin^n(18x)

dexar [7]

Let

$I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx$
For demonstration on how to tackle this sort of problem, I'll only work through the case where $n$ is odd. We can write

$\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx$
$\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$

For the remaining integral, we can integrate by parts, taking

$u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx$ $\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax$

$\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx$

For this next integral, we rewrite the integrand

$\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax$
$\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)$

So putting everything together, we found

$I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$
$I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)$
$I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$
$\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$

$\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax$

7 0

3 years ago

Y-2 = -3(x + 6), write the equation of the line in slope intercept form.

lisov135 [29]

Answer:

y = -3x - 16

Step-by-step explanation:

Point-slope form: y - y1 = m(x - x1)

Given: y - 2 = -3(x + 6)

Slope-intercept form: y = mx + b

To write the equation in slope-intercept form, we need to know the slope(m) and the y-intercept(b). By looking at the given equation, we can see that the slope is -3. We are also given the value of one point: (-6, 2).

To find the y-intercept, input the given values of the slope and the point into the equation format and solve for b:

y = mx + b

2 = -3(-6) + b

2 = 18 + b

Subtract 18 from both sides of the equation to isolate b:

2 - 18 = 18 - 18 + b

-16 = b

The y-intercept is -16.

Now that we know the slope and the y-intercept, we can write the equation:

y = -3x - 16

8 0

3 years ago