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2 years ago

PLZ help ASAP......Find the area of the following figure.....

Mathematics

2 answers:

Lady_Fox [76]2 years ago

5 0

Answer:

196 cm²

Step-by-step explanation:

[Area of Triangle BCD]

A = $\frac{1}{2}$ bh = $\frac{1}{2}$ * 28 * 6 = 84 cm²

[Area of Triangle ABD]

A = $\frac{1}{2}$ bh = $\frac{1}{2}$ * 28 * 8 = 112 cm²

dedylja [7]2 years ago

5 0

Hi Friend Your answer is in photo check plzz

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PLZ HELP ME

NeTakaya

Answer:

see below

Step-by-step explanation:

The shading is below the line in both cases. The dashed line has negative slope, so the inequality associated with it is ...

y < -2x +4 . . . . . . note that the < symbol has no "or equal to"

The solid line has positive slope, so its inequality is ...

y ≤ 2x +4

Both these inequalities are found in choice B.

6 0

3 years ago

How to find the hypotenuse

masya89 [10]

Answer:

use Pythagoras theorem to find hypotenuse in right angle triangle

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3 years ago

What are congruence of a triangle

Katen [24]

Answer:

it is when two triangles are similar to each other in every way.

Mainly there are 4 TYPES OF CONGRUENCY.(A.A.S, S.S.S, R.H.S, S.A.S )

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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

On a number line, what is the distance between -0.6 to -0.4?

Mila [183]

Answer is B
-0.6- (-0.4)= -0.6+0.4= -0.2

4 0

3 years ago

Read 2 more answers

PLZ help ASAP......Find the area of the following figure.....​

PLZ help ASAP......Find the area of the following figure.....