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2 years ago

Please put details! I need it!

Mathematics

1 answer:

Kay [80]2 years ago

3 0

Answer:

a. the points will be in quadrants q and s

b. the difference between the coordinates of the starting point and the checkpoint is that the starting point is negative on the 5 axis while the checkpoint is positive. the number on the y axis for the staring point and the checkpoint is the same.

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Guys please answer question #17 and 18 below please and thank you. Sorry about this photo has been rotated, I'm gonna try my bes

Contact [7]

Answer:

i think its 3

Step-by-step explanation:

3 0

2 years ago

What is the solution to the linear equation? Show your work.<br> -12 + 3b - 1 = -5 - b

butalik [34]

Answer:

b = 2

Step-by-step explanation:

Original Equation:

-12 + 3b - 1 = -5 - b

Add b to both sides

-12 + 4b - 1 = -5

Add 12 to both sides

4b - 1 = 7

Add 1 to both sides

4b = 8

Divide both sides by 4

b = 2

Hope this is what you need :)

3 0

3 years ago

Read 2 more answers

Simplify 1/2+2/3-1/4

avanturin [10]

Answer:

11/12

Step-by-step explanation:

5 0

2 years ago

PLEASE HELP ME I'M GIVING 20PTS AND MARKING BRAINLIEST!!!!

pishuonlain [190]

Using a trigonometric identity, it is found that the values of the cosine and the tangent of the angle are given by:

$\cos{\theta} = \pm \frac{2\sqrt{2}}{3}$

$\tan{\theta} = \pm \frac{\sqrt{2}}{4}$

<h3>What is the trigonometric identity using in this problem?</h3>

The identity that relates the sine squared and the cosine squared of the angle, as follows:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

In this problem, we have that the sine is given by:

$\sin{\theta} = \frac{1}{3}$

Hence, applying the identity, the cosine is given as follows:

$\cos^2{\theta} = 1 - \sin^2{\theta}$

$\cos^2{\theta} = 1 - \left(\frac{1}{3}\right)^2$

$\cos^2{\theta} = 1 - \frac{1}{9}$

$\cos^2{\theta} = \frac{8}{9}$

$\cos{\theta} = \pm \sqrt{\frac{8}{9}}$

$\cos{\theta} = \pm \frac{2\sqrt{2}}{3}$

The tangent is given by the sine divided by the cosine, hence:

$\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$

$\tan{\theta} = \frac{\frac{1}{3}}{\pm \frac{2\sqrt{2}}{3}}$

$\tan{\theta} = \pm \frac{1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$\tan{\theta} = \pm \frac{\sqrt{2}}{4}$

More can be learned about trigonometric identities at brainly.com/question/24496175

#SPJ1

5 0

1 year ago

Please answer i really wanna pass (;´༎ຶٹ༎ຶ`)

motikmotik

The answer is 2/3 shirt per hour.

4 0

3 years ago

Read 2 more answers