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4 years ago

If two parallel lines are cut by a transversal, then the two pairs of same-side interior angles are _________.

Mathematics

1 answer:

Andreas93 [3]4 years ago

5 0

Then they are supplementary angles which mean both of those angles add up to 180°

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-27=-3p+15-3p solve the equation and show your work

Gnesinka [82]

-27 = -3p + 15 - 3p

Combine like terms.
-3p -3p = (-3-3)p = -6p

-27 = -6p + 15

Subtract 15 from each side to get the variable term on its own.

-27 - 15 = -6p + 15 - 15

-42 = -6p

Divide each side by -6 to find p.

p = -42 ÷ -6 = 7

6 0

3 years ago

-10n+50=-7(4n+8)-2 no solution many solution

ziro4ka [17]

Answer:

n=-6

Step-by-step explanation:

distribute -7 into (4n+8)

-10n+50=-28n-56-2

combine like terms

-10n+50=-28n-58

add -28n on both sides

-10n+28n+50=-58

combine like terms again

18n+50=-58

subtract 50 on both sides

18n=-108

divide both sides by 18

$\frac{18x}{18} =\frac{-108}{18}$

x=-6

hope this helps

8 0

4 years ago

Which of the following functions are solutions of the differential equation y'' + y = 3 sin x? (select all that apply)

DiKsa [7]

Answer:

Step-by-step explanation:

We must compute the derivatives and check if the equation is satisfied.

A. $y'=(3\sin x)'=3\cos x$ . Differentiate again to get $y''=(3\cos x)'=-3\sin x$ , then $y''+y=-3\sin x+3\sin x=0\neq 3\sin x$ so this choice of y doesn't solve the equation.

B. $y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x$ and $y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x$ , then $y''+y=10\sin x+6\cos x\neq 3\sin x$ so y is not a solution

C. $y'=\frac{-3}{2}\cos x+\frac{3}{2}x\sin x$ hence $y''=\frac{3}{2}\sin x+\frac{3}{2}\sin x+\frac{3}{2}x\cos x=3\sin x+\frac{3}{2}x\cos x$ . Then $y''+y=3\sin x+\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\sin x$ so y is a solution.

D. $y'=-3\sin x$ and $y''=-3\cos x$ , then $y''+y=0$ thus y isn't a solution

E. $y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x$ hence $y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x$ . Then $y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x$ then y is not a solution.

3 0

3 years ago

-2 = g - 9 solve equation

jekas [21]

OK. Hold on tight. Here we go.
Remember ... the "solution" is the number that 'g' must be
in order to make the equation a true statement.

The original equation: You said that -2 = g - 9

Now, let's add 9 to each side: 7 = g .

8 0

3 years ago

Read 2 more answers

ASAP The ratio of the width to the length of a ratio is1:3 which of the following ratio represents a rectangle with an equivalen

Soloha48 [4]

Take whatever side lengths you have and divide that number by 1/3 in your calculator and what number is shown after doing this is the answer

4 0

3 years ago