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3 years ago

13

Determine if these values form a right triangle. E

1 answer:

LekaFEV [45]3 years ago

4 0

Answer:

They do not.

Step-by-step explanation:

a² + b² = c² c is the hypotenuse

9² + 12² = 17²

81 + 144 = 289

225 ≠ 289

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What is the simplified form of the quantity 9 x squared minus 25 over the quantity 3 x plus 5 ? (1 point) 3x + 5, with the restr

Solnce55 [7]

If so, start by factoring the numerator. Then cancel any common term in the numerator and denominator. If you end up with no denominator, then use the original denominator. Set it equal to zero and solve for x. That value of x will be the restriction.

7 0

3 years ago

What is the sum of the roots of the quadratic equation x²+6x-14=0?

aivan3 [116]

Answer:

<h3>Sum of Roots =<u> (-8)</u><u> </u><u> </u></h3><h3>Product of Roots = <u> </u><u> </u><u>(-84)</u><u> </u><u> </u></h3>

Step-by-step explanation:

${x}^{2} + 6x - 14 = 0$

Roots :- 6 and -14

Sum of the roots :

[6 + (-14)]

= (-8)//

Product of the roots :

[(6)(-14)]

= (-84)//

8 0

3 years ago

Solve the equation for a.

katrin [286]

Answer:

$a= \frac{k}{9b+4}$

Step-by-step explanation:

In this question, you would solve for "a".

Solve:

K = 4a + 9ab

Since we have our "a" on the same side, we can factor it out from the variables:

K = a(9b + 4)

To get "a" by itself, we would have to divide both sides by 9b + 4:

K/9b+ 4 = a

Your answer would be K/9b+ 4 = a

It would look like this: $a= \frac{k}{9b+4}$

8 0

3 years ago

Read 2 more answers

Simplify each ratio. <br><br> 1) 40:15

Vilka [71]

Answer:

8:3

Step-by-step explanation:

,,,,lol hope it helps i don't know how to explain,,, I guess its just dividing you know like 15÷5= 3 so on

3 0

3 years ago

Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).

fomenos

Answer:

$\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1$

Step-by-step explanation:

The standard equation of a horizontal hyperbola with center (h,k) is

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$

The given hyperbola has vertices at (–10, 6) and (4, 6).

The length of its major axis is $2a=|4--10|$ .

$\implies 2a=|14|$

$\implies 2a=14$

$\implies a=7$

The center is the midpoint of the vertices (–10, 6) and (4, 6).

The center is $(\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)$

We need to use the relation $a^2+b^2=c^2$ to find $b^2$ .

The c-value is the distance from the center (-3,6) to one of the foci (6,6)

$c=|6--3|=9$

$\implies 7^2+b^2=9^2$

$\implies b^2=9^2-7^2$

$\implies b^2=81-49$

$\implies b^2=32$

We substitute these values into the standard equation of the hyperbola to obtain:

$\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1$

$\frac{(x+3)^2}{49} -\frac{(y-6)^2}{32}=1$

7 0

3 years ago