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3 years ago

Triangle ABC is isosceles with AB = AC.

Mathematics

1 answer:

SVEN [57.7K]3 years ago

8 0

Answer:

area of traingle=1/2(sin130×AB×AC)

85×3=sin130×AB×AB

255=76×AB²

255/76=AB²

√332.88=AB

AB=18.24

AC=18.24

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Classify the conic section and write the equation in standard form:

Karo-lina-s [1.5K]

Answer:

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Step-by-step explanation:

6 0

3 years ago

10. Emmilynn gets paid $975 each month. If she opens a savings account with a $100 check she

blondinia [14]

Answer:

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5 0

3 years ago

HELLO. I NEED HELP ON A HANDFUL OF QUESTIONS. PLEASE HELP.

galben [10]

Question 7: $3945

78,900÷100×5= 3,945

or 789,00×0.05= $3,945

Question 8: I think it may be $675.46 ( but I'm not sure)

$1010 is her annual insurance premium so to find one month we need to divide $1010 by 12

$1010÷12= $84.20

also her annual real estate tax is $938 and to find one month we need to divide this by 12 as well.

$938÷12=$78.20

and in order to find out her combined monthly payment we need to add them all together

$513.12+$84.16+$78.16= $675.46

Question 9: is False

Question 10: I don't know

Hope this helps

5 0

3 years ago

8. Find an equation in standard form for the ellipse with the vertical major axis of length 16 and minor axis of length 4

MA_775_DIABLO [31]

Hello,

If center is (0,0)

x²/a²+y²/b²=1 for a half horizontal axis of a, and a half vertical axis of b

Here

a=2 ==> a²=4

b=8 ==> b²=64

$\boxed{\dfrac{x^2}{4}+ \dfrac{y^2}{64}=1}\\$

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=5%28sin%28t%29%20%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%2Bycos%28t%29%29%29%3Dcos%28t%29%20%20%28sin%28

Nataliya [291]

Assuming you mean

$5\sin t\dfrac{\mathrm dy}{\mathrm dt}+5y\cos t=\cos t\sin^2t$

This ODE is linear in $y$ , and you can already contract the left hand side as the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[5y\sin t\right]=\cos t\sin^2t$

Integrating both sides with respect to $t$ yields

$5y\sin t=\displaystyle\int\cos t\sin^2t\,\mathrm dt$
$5y\sin t=\dfrac13\sin^3t+C$
$y=\dfrac1{15}\sin^2t+C\csc t$

Given that $y\left(\dfrac\pi2\right)=9$ , we have

$9=\dfrac1{15}\sin^2\dfrac\pi2+C\csc\dfrac\pi2$
$9=\dfrac1{15}+C$
$C=\dfrac{134}{15}$

so that the particular solution over the interval is

$y=\dfrac1{15}\sin^2t+\dfrac{134}{15}\csc t$

6 0

3 years ago