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4 years ago

10

Answer thanks, show how you got the answer.

1 answer:

levacccp [35]4 years ago

5 0

Answer:

Y = 4X - 6

Step-by-step explanation:

Slope of linear equation = change in Y / change in X

As Y is changing +4 units, with change in X +1 units. [ -6 - (-2) ] / [ 1 - 0 ]

So, slope = 4

Y = A + BX

Y = A + 4X , as B = slope & A = intercept

Putting values of X & Y (0,-6)

-6 = A + 4 (0)

A = -6

Hence, Equation : Y = 4X - 6

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Find the sum.<br> 2x<br> x2 - 6x + 9 + x2 + 2x - 15

galina1969 [7]

Answer:

$\large\boxed{x^2-6x+9+x^2+2x-15=2x^2-4x-6}$

Step-by-step explanation:

$x^2-6x+9+x^2+2x-15\qquad\text{combine like terms}\\\\=(x^2+x^2)+(-6x+2x)+(9-15)\\\\=2x^2-4x-6$

5 0

3 years ago

I’ll mark brainlest if you get it right

Makovka662 [10]

Answer:

47 degrees

Step-by-step explanation:

Quadrilaterals add up to 360

The two dashes and the dashes imply that angle V and angle X are congruent.

128 + 138 = 266

360 - 266 = 94

94 / 2 = 47

5 0

4 years ago

Find all points on the curve x=4cos(t),y=4sin(t) that have the slope of 12.

Marianna [84]

Answer:

$\left (-\dfrac{4}{\sqrt{5}},\dfrac{8}{\sqrt{5}}\right )\text{ and }\left (\dfrac{4}{\sqrt{5}},-\dfrac{8}{\sqrt{5}}\right )$ .

Step-by-step explanation:

We need to find all the points on the curve x=4cos(t),y=4sin(t) that have the slope of 1/2.

$x=4cos (t)$

$\dfrac{dx}{dt}=-4sin (t)$

$y=4sin (t)$

$\dfrac{dy}{dt}=4cos (t)$

Now,

$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}$

$\dfrac{dy}{dx}=4cos (t)\times \dfrac{1}{-4sin (t)}$

$\dfrac{dy}{dx}=-\cot t$

So, slope of the curve is $-\cot t$ .

$-\cot t=\dfrac{1}{2}$

$-\tan t=2$

$\tan t=-2$ ...(1)

Using $\sec^2t=1+\tan^2t$ , we get

$\sec^2t=1+(-2)^2$

$\sec^2t=1+4$

$\sec t=\pm \sqrt{5}$

$\cos t=\pm \dfrac{1}{\sqrt{5}}$

Now,

$\sin^2t=1-cos^2t$

$\sin t=\pm \sqrt{1-\dfrac{1}{5}}$

$\sin t=\pm \sqrt{\dfrac{4}{5}}$

$\sin t=\pm \dfrac{2}{\sqrt{5}}$

It equation (1), tan(t) is negative. So, sin and cos have different signs.

If $\sin t= \dfrac{2}{\sqrt{5}}$ , then $\cos t=- \dfrac{1}{\sqrt{5}}$ .

$x=4cos (t)=-\dfrac{4}{\sqrt{5}}$

$y=4sin (t)=\dfrac{8}{\sqrt{5}}$

If $\sin t=- \dfrac{2}{\sqrt{5}}$ , then $\cos t= \dfrac{1}{\sqrt{5}}$ .

$x=4cos (t)=\dfrac{4}{\sqrt{5}}$

$y=4sin (t)=-\dfrac{8}{\sqrt{5}}$

Therefore, the two points are $\left (-\dfrac{4}{\sqrt{5}},\dfrac{8}{\sqrt{5}}\right )\text{ and }\left (\dfrac{4}{\sqrt{5}},-\dfrac{8}{\sqrt{5}}\right )$ .

5 0

3 years ago

Find the measure of the side indicated by the variable of x. Round your answer to the nearest tenth.

Alex17521 [72]

Answer:

13.1

message me if you need more help

7 0

3 years ago

The slide at the playground has a height of 6 feet. The base of the slide measured on the ground is 8 feet. What is the length o

Damm [24]

Answer:

10 feet

Step-by-step explanation:

a²+b²+=c²

a=8ft

b=6ft

4 0

3 years ago