Find ST if S is between R and T, SR = 2x + 7, ST = x - 6, and RT = 100

1 answer:

7 0

Since SR+ST adds up to RT, we have 2x+7+x-6=100=3x+1.
Subtracting 1 from both sides, we get 3x=99

Dividing both sides by 3, we have x=33

You might be interested in

Which of the following are solutions to the equation below? Check all that apply. 9x^2-64=0

nikdorinn [45]

Answer:

E. -8/3

F. 8/3

Step-by-step explanation:

9x^2-64=0

Add 64 to each side

9x^2-64+64 = 0+64

9x^2 =64

Divide each side by 9

9x^2/9 = 64/9

x^2 = 64/9

Take the square root of each side

sqrt(x^2) = ±sqrt(64/9)

We know that sqrt(a/b) = sqrt(a)/sqrt(b)

x = ±sqrt(64)/sqrt(9)

x = ± 8/3

4 0

3 years ago

PLZZZ HELLP I'LL GIVE YOU POINTS AND MARK YOU AS BRAINLIEST

yanalaym [24]

You plug 11 in where x is.
r(11) = (11 + 1)(11 - 3)
= (12)(8)
= 96

3 0

3 years ago

Write an equation in standard form of the hyperbola described.

marishachu [46]

Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.

$\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill$

$\begin{cases} h=0\\ k=0\\ a=2\\ c=4 \end{cases}\implies \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{b^2} \\\\\\ c^2=a^2+b^2\implies 4^2=2^2+b^2\implies 16=4+b^2\implies \underline{12=b^2} \\\\\\ \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{12}\implies \boxed{\cfrac{x^2}{4}-\cfrac{y^2}{12}}$