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3 years ago

Suppose an isosceles triangle ABC has A = 45° and b = c = 4. What is the length of a2?

1 answer:

const2013 [10]3 years ago

4 0

You can use the Law of Cosines (a^2=b^2+c^2-2bc cosA)

a^2=4^2+4^2-2*4*4*cos45°

a^2=32-32cos45°

a^2=32(1-cos45°)

a=√(32(1-cos45°))

a≈3.06 units (to nearest hundredth of a unit)

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Circle A has an area of 452.16ft2. With a radius of 12 ft. If circle B has a radius of 17 ft, what is it’s area?

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Answer: 907.92 ft²

Step-by-step explanation:

1. To solve this problem you must apply the formula for calculate the area of a circle, which is shown below:

$A=r^{2}\pi$

Where r is the radius of the circle.

2. You know the radius of the circle B, therefore, when you susbtitute it into the formula, you obtain that the area of the circle B is:

$A_B=(17ft)^{2}\pi=907.92ft^{2}$

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13x=-21-6y and -8x=36+6y

IgorLugansk [536]

Answer:

$x=3\\y=-10$

Step-by-step explanation:

$13x=-21-6y\\-8x=36+6y$

Let's take one of the equations and solve for x. I'll take the first one.

$13x=-21-6y$

Divide by 13.

$x=\frac{-21-6y}{13}$

Now replace x in the second equation.

$-8x=36+6y\\-8(\frac{-21-6y}{13} )=36+6y$

Distribute -8

$\frac{168+48y}{13}=36+6y$

Break down the fraction.

$\frac{168}{13}+\frac{48}{13}y=36+6y$

Subtract $\frac{168}{13}$

$\frac{48}{13}y=36+6y-\frac{168}{13}$

Subtract 6y

$\frac{48}{13}y-6y=36-\frac{168}{13}$

Combine like terms;

$\frac{48-13*6}{13}y=\frac{36*13-168}{13}$

Solve;

$\frac{48-78}{13}y=\frac{468-168}{13}$

Keep solving;

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Multiply by the negative reciprocal of the fraction next to y.

$(-\frac{13}{30} )\frac{-30}{13}y=\frac{300}{13}(-\frac{13}{30} )$

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