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3 years ago

PLEASE HELP!!!!!

Mathematics

1 answer:

GrogVix [38]3 years ago

5 0

Answer:

Step-by-step explanation:

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If R is the midpoint of QS, QR= 8x-51 and RS = 3x-6, find QS.

sashaice [31]

QS would be 42. So the way i did it was I:

8x-51=3x-6

x=9

Plugged into both equations of QR and RS to then get the answer of 42

7 0

3 years ago

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Write the equation of a line that is perpendicular to y=3x-2 that passes trough the point (-9,5)

ElenaW [278]

$\huge\boxed{y-5&=-\frac{1}{3}(x+9)}$

First, we'll find the slope of the new line. The first line has a slope of $3$ . Take the negative reciprocal of this (Flip the numerator and denominator, then multiply by $-1$ ) to get $-\frac{1}{3}$ for the new slope.

Then, we'll use the point-slope form to make the new equation, where $m$ is the slope and $(x_1,y_1)$ is a point on the line:

$\begin{aligned}y-y_1&=m(x-x_1)\\y-5&=-\frac{1}{3}(x-(-9))\\y-5&=-\frac{1}{3}(x+9)\end{aligned}$

3 0

1 year ago

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Find the right quotient. 36m 5 n 5 ÷ (12m 3)

polet [3.4K]

ANSWER

The right quotient is

$3 {m}^{2} {n}^{5}$

EXPLANATION

The given expression is :

$\frac{36 {m}^{5} {n}^{5} }{12 {m}^{3} }$

$\frac{36 {m}^{5} {n}^{5} }{12 {m}^{3} } = \frac{36}{12} \times \frac{ {m}^{5} }{ {m}^{3}} \times {n}^{5}$

$\frac{36 {m}^{5} {n}^{5} }{12 {m}^{3} } = 3\times \frac{ {m}^{5} }{ {m}^{3}} \times {n}^{5}$

Recall that,

$\frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$
We apply this property to obtain;

$\frac{36 {m}^{5} {n}^{5} }{12 {m}^{3} } = 3 \times {m}^{5 - 3} \times {n}^{5}$

$\frac{36 {m}^{5} {n}^{5} }{12 {m}^{3} } = 3 {m}^{2} {n}^{5}$

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2 years ago

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Determine the length of the leg of a 45o – 45o – 90o triangle with a hypotenuse length of 15 inches. ( ANSWER NEEDS TO BE IN RED

SIZIF [17.4K]

Answer:

The lenghts of both legs: $\frac{15\sqrt{2}}{2}\ inches$

Step-by-step explanation:

By definition, when a triangle has angles that measures 45°, 45° and 90°, its legs are congruent.

Then, knowing the lenght of the hypotenuse, we can find the lenght (in inches) of any leg of the given triangle by applying the Trigonometric Identity $sin\alpha=\frac{opposite}{hypotenuse}$ :