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Luba_88 [7]
3 years ago
9

GI and JL are parallel lines. Which angles are supplementary angles?

Mathematics
1 answer:
solmaris [256]3 years ago
6 0

Answer:

The answer is A.

Step-by-step explanation:

A supplementary angles are angles that when adding both sides will sum up to 180 Degrees.

<JKH and <LKH will be the correct answer because they add up tp 180 degrees.

Star Me!!!

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Consider the graph shown, which function eventually exceeds the other functions? A) y = 2x B) y = x2 C) y = 2x D) cannot be dete
nirvana33 [79]

Answer:

All of these are the same. I would say D as theya re the same.

7 0
3 years ago
Solve the system by the substitution method -4x + 3y = -7 y = 2x - 5
bagirrra123 [75]

Answer: x=4 and y= 3


Step-by-step explanation: Solve for the first variable in one of the equations, then substitute the result into the other equation


8 0
3 years ago
!!! WILL GIVE BRAINLIEST !! PLS ANSWER !!!
Brut [27]

Answer:

Yes, ASA

Step-by-step explanation:

It is angle- side- angle because of the HLK< it is ASSA but since the SS are the same, its ASA.

Hope this helps!

4 0
3 years ago
What is the constant term of the quotient when 2x^3-3x^2 + 4x-2 is divided by x^2-x+2?
AlladinOne [14]

Answer:

  -1

Step-by-step explanation:

See the attachment for the polynomial long division. The constant in the quotient is -1.

_____

Here, there is a remainder of -x. If there were no remainder the constant in the quotient is the ratio of the constant in the dividend to the constant in the divisor: -2/2 = -1.

That could be a first guess in a "guess and check" solution approach.

<em>Guess</em>: first term of binomial quotient is (2x^3)/x^2 = 2x; last term of binomial quotient is -2/2 = -1. So, the quotient is guessed to be (2x -1).

<em>Check</em>: (2x -1)(x^2 -x +2) = 2x^3 -3x^2 +5x -2

Subtracting this from the actual dividend gives a remainder of -x. This has a lower degree than the divisor, so no further adjustment of the quotient is required.

6 0
3 years ago
Integrate Sec (4x - 1) Tan (4x - 1) Dx ​
Delicious77 [7]

\bf \displaystyle\int~sec(4x-1)tan(4x-1)dx \\\\[-0.35em] ~\dotfill\\\\ sec(4x-1)tan(4x-1)\implies \cfrac{1}{cos(4x-1)}\cdot \cfrac{sin(4x-1)}{cos(4x-1)}\implies \cfrac{sin(4x-1)}{cos^2(4x-1)} \\\\[-0.35em] ~\dotfill\\\\ \displaystyle\int~\cfrac{sin(4x-1)}{cos^2(4x-1)}dx \\\\[-0.35em] ~\dotfill\\\\ u=cos(4x-1)\implies \cfrac{du}{dx}=-sin(4x-1)\cdot 4\implies \cfrac{du}{-4sin(4x-1)}=dx \\\\[-0.35em] ~\dotfill

\bf \displaystyle\int~\cfrac{~~\begin{matrix} sin(4x-1) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{u^2}\cdot \cfrac{du}{-4~~\begin{matrix} sin(4x-1) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies -\cfrac{1}{4}\int\cfrac{1}{u^2}du\implies -\cfrac{1}{4}\int u^{-2}du \\\\\\ -\cfrac{1}{4}\cdot \cfrac{u^{-2+1}}{-1}\implies \cfrac{1}{4}\cdot u^{-1}\implies \cfrac{1}{4u}\implies \cfrac{1}{4cos(4x+1)}+C

5 0
4 years ago
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