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4 years ago

Which equation below represents a horizontal line?

Mathematics

2 answers:

Debora [2.8K]4 years ago

8 0

Hello :
answer A) y = 4 because the slope = 0

jeka57 [31]4 years ago

7 0

Answer:A) y = 4

Step-by-step explanation:

The slope of any horizontal line is zero, 0 .The equation of any horizontal line is of the form = b, where b is the y-coordinate of the y-intercept .

Among all the options given The equation y=4 represent a horizontal line .

The equations y=x-0 ,y=x+0 represent the same line y=x and are not horizontal lines. x=4 is a vertical line.

Option A y=4 represent a horizontal line.

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(50-(x*50))+(40-(y*40))+20= discounted with 2 coupons

110-(z*110)= the price with the total coupon.

I believe these are the correct answers

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3 years ago

Read 2 more answers

A is C's son. C and A are sisters. Z is Q's mother. P is son of Z. Then

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C) q is a grandfather

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3 years ago

What is the best estimate for 56.23 - 45.9? A. 9 B. 10 C. 11 D. 15

DIA [1.3K]

Answer:

10 Letter B

Step-by-step explanation:

56.23 rounds down to 56 and 45.9 rounds up to 46. So 56 - 46 = 10

5 0

3 years ago

Save me the headache

maxonik [38]

$(9\sin2x+9\cos2x)^2=81$

Taking the square root of both sides gives two possible cases,

$9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1$

$9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1$

Recall that

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

If $\alpha=2x$ and $\beta=\dfrac\pi4$ , we have

$\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}$

so in the equations above, we can write

$\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1$

Then in the first case,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi$

(where $n$ is any integer)

$\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi$

$\implies x=n\pi\text{ or }\dfrac\pi4+n\pi$

and in the second,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi$

$\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi$

$\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi$

Then the solutions that fall in the interval $[0,2\pi)$ are

$x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4$

5 0

3 years ago

Read 2 more answers

After unloading 96 pounds of mulch from the truck, there were still 24 pounds

Dennis_Churaev [7]

Answer:

120

Step-by-step explanation:

To find the answer you just have to add 24 and 96 together to get 120.

4 0

3 years ago