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Mathematics

1 answer:

ValentinkaMS [17]3 years ago

7 0

Step-by-step explanation:

hope it is helpful to you

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Angle $eab$ is a right angle, and $be = 9$ units. what is the number of square units in the sum of the areas of the two squares

Alla [95]

Let the measure of side AB be x, then, the measue of side AE is given by

$AE=\sqrt{9^2-x^2}$ .

Now, ABCD is a square of size x, thus the area of square ABCD is given by

$Area=x^2$

Also, AEFG is a square of size $\sqrt{9^2-x^2}$ , thus, the area of square AEFG is given by

$Area=\left(\sqrt{9^2-x^2}\right)^2=9^2-x^2=81-x^2$

The sum of the areas of the two squares ABCD and AEFG is given by

$x^2+81-x^2=81$

Therefore, the number of square units in the sum of the areas of the two squares ABCD and AEFG is 81 square units.

8 0

3 years ago

F(x)=12x2-4x-8. G(x)=11x-6

AlekseyPX

Answer: Second option.

Step-by-step explanation:

Given the functions f(x) and g(x):

$f(x)=12x^2-4x-8\\\\g(x)=11x-6$