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3 years ago

Which statement about numbers on the horizontal number line is true?

2 answers:

8 0

<span>C.) On the horizontal number line, -0.75 is located to the left of -0.4

[ It is because, When you go from left to right,number decreased upto zero, and then increases with positive sign,so more negative number would be in left and more positive in the right]

Hope this helps!</span>

sattari [20]3 years ago

6 0

On a horizontal number line, -ve is on left, 0 in middle n +ve on right

so ans is <span>C.) On the horizontal number line, -0.75 is located to the left of -0.4.</span>

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If Cyrus knows that △ABC∼△EDC and AB¯¯¯¯¯¯¯¯∥ED¯¯¯¯¯¯¯¯, how can he prove that line m passing through AB¯¯¯¯¯¯¯¯ has the same sl

Illusion [34]

As AB and ED are parallel, they have the same slope. This means that since lines that are colllinear with parallel segments are parallel, lines l and m are parallel.

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1 year ago

Find the quotient. -22s^4t -8st /2st

kicyunya [14]

Answer:

-11s^3 -4

Step-by-step explanation:

( -22s^4t -8st) /2st

Separate into parts

-22s^4t / 2st -8st /2st

-11 s^3 -4

8 0

3 years ago

Cosθ=−2√3 , where π≤θ≤3π2 .

Alex787 [66]

Answer:

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we know that

Applying the trigonometric identity

$sin^2(\theta)+ cos^2(\theta)=1$

we have

$cos(\theta)=-\frac{\sqrt{2}}{3}$

substitute

$sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1$

$sin^2(\theta)+ \frac{2}{9}=1$

$sin^2(\theta)=1- \frac{2}{9}$

$sin^2(\theta)= \frac{7}{9}$

$sin(\theta)=\pm\frac{\sqrt{7}}{3}$

Remember that

π≤θ≤3π/2

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

$sin(\theta)=-\frac{\sqrt{7}}{3}$

step 2

Find the sec(β)

Applying the trigonometric identity

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

we have

$tan(\beta)=\frac{4}{3}$

substitute

$(\frac{4}{3})^2+1= sec^2(\beta)$

$\frac{16}{9}+1= sec^2(\beta)$

$sec^2(\beta)=\frac{25}{9}$

$sec(\beta)=\pm\frac{5}{3}$

we know

0≤β≤π/2 ----> II Quadrant

sec(β), sin(β) and cos(β) are positive

$sec(\beta)=\frac{5}{3}$

Remember that

$sec(\beta)=\frac{1}{cos(\beta)}$

therefore

$cos(\beta)=\frac{3}{5}$

step 3

Find the sin(β)

we know that

$tan(\beta)=\frac{sin(\beta)}{cos(\beta)}$

we have

$tan(\beta)=\frac{4}{3}$

$cos(\beta)=\frac{3}{5}$

substitute

$(4/3)=\frac{sin(\beta)}{(3/5)}$

therefore

$sin(\beta)=\frac{4}{5}$

step 4

Find sin(θ+β)

we know that

$sin(A + B) = sin A cos B + cos A sin B$

In this problem

$sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)$

we have

$sin(\theta)=-\frac{\sqrt{7}}{3}$

$cos(\theta)=-\frac{\sqrt{2}}{3}$

$sin(\beta)=\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute the given values in the formula

$sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})$

$sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})$

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

8 0

3 years ago

HELP WILL MARK BRAINIEST!!!

maria [59]

Answer:

23 feet

Step-by-step explanation:

The absolute value of -14 is fourteen.

The absolute value of 23 is twenty-three

23>14

4 0

2 years ago

I need help, please!!! I got the answer but I need to show work and that I got that answer.

Rainbow [258]

Step-by-step explanation:

Area of each paralellogram = b * h = 12 * 20 = 240 square inches

Area of 9 paralellograms = 9 * 240 = 2160 square inches

5 0

2 years ago

Read 2 more answers