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Kipish [7]
2 years ago
14

A triangle has an area of 24 square cm and a base of 4 cm. Which is the

Mathematics
1 answer:
devlian [24]2 years ago
3 0
The answer is C. So if u plug it in the equation should look something like this: 4*12 / 2
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Draw the image of B(7,-4) under a reflection over x=2
lesantik [10]

Answer:

Put the green dot at the coordinates (-3, -4). Find -3 on the x-axis and go 4 down.

Step-by-step explanation:

Just move the blue dot five to the left, then five to the left again.

3 0
2 years ago
Can i get some help on my test please?
Veseljchak [2.6K]

Answer:

Not proportional.

Step-by-step explanation:

The butterfly method multiples it like a butterfly. Ex. 5 times 3 is 15. 8 times 2 is  16. Therefore, not proportional.

5 0
2 years ago
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Find sin A for the triangle below. Give the exact value as an expression and an approximation to the nearest ten-thousandth. Not
Feliz [49]

the exact value of sin A  to the nearest ten- thousandth is 0. 6625

<h3>Using the pythagorean theorem</h3>

a² + b² = c²

The opposite side is unknown, so use the pythagorean theorem to find it

c = hypotenuse = 4

a= opposite site = ?

b= adjacent side = 3

Substitute into the formula

4² = a² + 3²

16 = a² + 9

a² = 16 -9 = 7

Find the square root

a =√7 = 2. 65

To find Sin A, use

Sin A = opposite side ÷ hypotenuse

Sin A = 2. 65 ÷ 4 = 0. 6625

Thus,  the value of sin A is 0. 6625

Learn more about pythagorean theorem here:

brainly.com/question/654982

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7 0
2 years ago
Solution for dy/dx+xsin 2 y=x^3 cos^2y
vichka [17]
Rearrange the ODE as

\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y
\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3

Take u=\tan y, so that \dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}.

Supposing that |y|, we have \tan^{-1}u=y, from which it follows that

\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}
\sec^2y=1+\tan^2y=1+u^2

So we can write the ODE as

\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3

which is linear in u. Multiplying both sides by e^{x^2}, we have

e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}
\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}

Integrate both sides with respect to x:

\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx
e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx

Substitute t=x^2, so that \mathrm dt=2x\,\mathrm dx. Then

\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt

Integrate the right hand side by parts using

f=t\implies\mathrm df=\mathrm dt
\mathrm dg=e^t\,\mathrm dt\implies g=e^t
\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)

You should end up with

e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C
u=\dfrac{x^2-1}2+Ce^{-x^2}
\tan y=\dfrac{x^2-1}2+Ce^{-x^2}

and provided that we restrict |y|, we can write

y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)
5 0
3 years ago
Which of the following mortgage options will not have a PMI requirement?
uranmaximum [27]
I think the answer c
6 0
3 years ago
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