7. If the longer leg of a right triangle is twice as long as the shorter leg, what is the ratio of the measure if the larger acute angle to the smaller acute angle?

The ratio is 2.39, which means that the larger acute angle is 2.39 times the smaller acute angle.

Step-by-step explanation:

I suppose that the "legs" of a triangle rectangle are the cathati.

if L is the length of the shorter leg, 2*L is the length of the longest leg.

Now you can remember the relation:

Tan(a) = (opposite cathetus)/(adjacent cathetus)

Then there is one acute angle calculated as:

Tan(θ)  =  (shorter leg)/(longer leg)

Tan(φ) = (longer leg)/(shorter leg)

And we want to find the ratio between the measure of the larger acute angle and the smaller acute angle.

Then we need to find θ and φ.

Tan(θ) = L/(2*L)

Tan(θ) = 1/2

θ = Atan(1/2) = 26.57°

Tan(φ) = (2*L)/L

Tan(φ) = 2

φ = Atan(2) = 63.43°

Then the ratio between the larger acute angle and the smaller acute angle is:

R = (63.43°)/(26.57°) = 2.39

This means that the larger acute angle is 2.39 times the smaller acute angle.

when studying polynomials, you often hear the terms zeros, roots, factors and x x the multiplicity of a zero is important because it tells us how the graph of the let's look at the graph of a function that has the same zeros

step-by-step explanation:

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2