Find the solution(s) to the system of equations. Select all that apply.

Mathematics

1 answer:

pickupchik [31]3 years ago

7 0

Answer:

A. (4, 5) and C. (0, -3).

Step-by-step explanation:

A. 5 = 4 * 4 - 2 * 4 - 3

5 = 16 - 8 - 3

5 = 8 - 3

5 = 5

5 = 2 * 4 - 3

5 = 8 - 3

5 = 5

A. (4, 5) is a solution.

B. 0 = 1 + 2 - 3

0 = 0

0 = -2 - 3

0 IS NOT EQUAL TO -5

B. (-1, 0) is not a solution.

C. -3 = 0 - 0 -3

-3 = -3

-3 = 0 - 3

-3 = -3

C. (0, -3) is a solution.

D. 0 = 9 - 6 - 3

0 = 9 - 9

0 = 0

0 = 6 - 3

0 IS NOT EQUAL TO 3.

D. (3, 0) is not a solution.

So, your answers are A. (4, 5) and C. (0, -3).

Hope this helps!

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If the given values in a right triangle are cos A =

Vlad1618 [11]

See the picture attached to better understand the problem

we know that
in the right triangle ABC
cos A=AC/AB
cos A=1/3
so
1/3=AC/AB----->AB=3*AC-----> square----> AB²=9*AC²----> equation 1

applying the Pythagoras Theorem
BC²+AC²=AB²-----> 2²+AC²=AB²---> 4+AC²=AB²----> equation 2

substitute equation 1 in equation 2
4+AC²=9*AC²----> 8*AC²=4----> AC²=1/2----> AC=√2/2
so
AB²=9*AC²----> AB²=9*(√2/2)²----> AB=(3√2)/2

the answer is
the hypotenuse is (3√2)/2