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3 years ago

What is the measure of A using the triangle abc

Mathematics

1 answer:

djyliett [7]3 years ago

5 0

Do sin(90)*13/15 which gives you 0.87 then do inverse of sin-1(0.87) which angle A = 60.45 degrees

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3^x-3^(x-2)=A*3^x <br><br> solve for A

ale4655 [162]

Answer:

$a=\frac{8}{9}$

Step-by-step explanation:

$3^x-3x^{x-2} =a3^x$

Divide both sides by $3^x$ :

$3^{x-x}-3^{x-x-2}=a$

Simplify:

$3^0-3^{-2}=a\\a=1-\frac{1}{9} \\a=\frac{8}{9}$

3 0

3 years ago

Pls explain below :)

matrenka [14]

Answer:

$9 units^{2}$ aka 9 units

Step-by-step explanation:

Its pretty simple so what u do is count the sqaures then multiply them and ill do it for you

H=3

B=6

and the formula for a triangle is A= $\frac{b*h}{2}$ so its 3x6/2=9

The answer is 9

5 0

3 years ago

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Help asap please:<br><br> Solve x-7y=16 for y

Pepsi [2]

Answer:

y = (x-16)/7

Step-by-step explanation:

x - 7y = 16

~Subtract x to both sides

-7y = 16 - x

~Divide -7 to everything

y = (16-x)/7

Best of Luck!

5 0

3 years ago

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Angle ABC is formed by segments AB and BC on the following coordinate grid: A coordinate grid is shown from positive 6 to negati

Nadusha1986 [10]

Answer:

B) m∠ A'B'C' = m∠ABC

Step-by-step explanation:

A rotation does not change angle measure or side length. It preserves congruence.

This means that the image angle, A'B'C', will be congruent to the pre-image angle, ABC.

5 0

3 years ago

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Find k so that the distance from (–1, 1) to (2, k) is 5 units. k= k= *there are two solutions for 2*

dalvyx [7]

Answer:

k = -3

k =5

Step-by-step explanation:

$d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\d = 5\\(-1,1) =(x_1,y_1)\\(2,k)=(x_2,y_2)\\$

$5=\sqrt{\left(2-\left(-1\right)\right)^2+\left(k-1\right)^2}\\\\\mathrm{Square\:both\:sides}:\quad 25=k^2-2k+10\\25=k^2-2k+10\\\\\mathrm{Solve\:}\:25=k^2-2k+10:\\k^2-2k+10=25\\\\\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\k^2-2k+10-25=25-25\\k^2-2k-15=0\\\\\mathrm{Solve\:by\:factoring}\\\\\mathrm{Factor\:}k^2-2k-15:\quad \left(k+3\right)\left(k-5\right)\\\mathrm{Solve\:}\:k+3=0:\quad k=-3\\$

$\mathrm{Solve\:}\:k-5=0:\quad k=5\\\\k =5 , k=-3$

7 0

3 years ago

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