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3 years ago

Help please and I will help u out with anything even boyfriend trouble or girlfriend trouble

Mathematics

2 answers:

scoundrel [369]3 years ago

4 0

You should get -15+(-5r)

ohaa [14]3 years ago

4 0

You should get 5+(-5r) cause -5+10=5

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Rewrite the slope-intercept form equation into standard form.<br> y = 5/2

Dahasolnce [82]

2y = 5 hope this helps :)

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3 years ago

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svlad2 [7]

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3 years ago

Assume that the terminal side of thetaθ passes through the point (negative 12 comma 5 )(−12,5) and find the values of trigonomet

zmey [24]

Answer:

$\sin \theta = \dfrac{5}{13}$ and $\sec \theta = -\dfrac{13}{12}$

Step-by-step explanation:

Assume that the terminal side of thetaθ passes through the point (−12,5).

In ordered pair (-12,5), x-intercept is negative and y-intercept is positive. It means the point lies in 2nd quadrant.

Using Pythagoras theorem:

$hypotenuse^2=perpendicular^2+base^2$

$hypotenuse^2=(5)^2+(12)^2$

$hypotenuse^2=25+144$

$hypotenuse^2=169$

Taking square root on both sides.

$hypotenuse=13$

In a right angled triangle

$\sin \theta = \dfrac{opposite}{hypotenuse}$

$\sin \theta = \dfrac{5}{13}$

$\sec \theta = \dfrac{hypotenuse}{adjacent}$

$\sec \theta = \dfrac{13}{12}$

In second quadrant only sine and cosecant are positive.

$\sin \theta = \dfrac{5}{13}$ and $\sec \theta = -\dfrac{13}{12}$

6 0

3 years ago

-18/-32/41/8/-11 from least to greatest

kifflom [539]

Answer:

$-32, -18, -11, 8, 41$

Step-by-step explanation:

Positive integers have values greater than zero. Negative integers have values less than zero.

$-32, -18, -11, 8, 41$

5 0

3 years ago

An equation is shown below:

marin [14]

10x -15 = 5
10x = 20
x = 2

This equation has one solution.

The solution would be x = 2

3 0

3 years ago