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3 years ago

Inequality stuff, please help!

Mathematics

1 answer:

aleksandrvk [35]3 years ago

4 0

Answer:

A. the ordered pair on he graph is based on the y-intercept which is: .0 1/2

B.The equantity is : Y=.1x+2/3

Step-by-step explanation:

Hopefully this helped you!

Have a wonderful day! :)

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3x + 1 − 4x3 + 6x6 −2x2 standard form

Molodets [167]

7 0

3 years ago

Choose the best score on a quiz. 24 points out of 42 24 points out of 42 75 points out of 135 75 points out of 135 150 points ou

stellarik [79]

Answer:

150/250

Step-by-step explanation:

24/42, 75/135, 150/250, and 75/150 I believe is what you're asking. We know 75/135 is better than 75/150 so 75/150 is out of the equation. If we multiple 75/135 by 2 we get 150/270 which is more than 150/250, so 75/135 is out of the equation. Now to compare 24/42 and 150/250 we can divide their fractions, if we divide 24/42 we get 0.57, so 57%. If we did 150/250 we get 0.6 or 60%. So 150/250 is the best score.

8 0

2 years ago

Read 2 more answers

I will give brainliest please hurry

Eduardwww [97]

Answer:

Step-by-step explanation:

300, since the line inside the box is on 300

Hope that helps!

6 0

2 years ago

Read 2 more answers

Tim spent $37.45 on groceries. On the way home, he wants to stop and pay cash for gas at $3.35 per gallon. If he started with $6

dalvyx [7]

Answer:

6.7

Step-by-step explanation:

60$-37.45=22.55

22.55/3.35=6.731

Which rounds to 6.7

5 0

3 years ago

Rewrite the expression in terms of the given function 1/1-sinx - sinx/1+sinx

Feliz [49]

Your question seems a bit incomplete, but for starters you can write

$\dfrac1{1-\sin x}-\dfrac{\sin x}{1+\sin x}=\dfrac{1+\sin x}{(1-\sin x)(1+\sin x)}-\dfrac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)}=\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}$

Expanding where necessary, recalling that $(1-\sin x)(1+\sin x)=1-\sin^2x=\cos^2x$ , you have

$\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}=\dfrac{1+\sin x-\sin x+\sin^2x}{\cos^2x}=\dfrac{1+\sin^2x}{\cos^2x}$

and you can stop there, or continue to rewrite in terms of the reciprocal functions,

$\dfrac{1+\sin^2x}{\cos^2x}=\sec^2x+\tan^2x$

Now, since $1+\tan^2x=\sec^2x$ , the final form could also take

$\sec^2x+\tan^2x=\sec^2x+(\sec^2x-1)=2\sec^2x-1$

or

$\sec^2x+\tan^2x=(1+\tan^2x)+\tan^2x=1+2\tan^2x$

7 0

3 years ago