All categories

3 years ago

Find the slope of the line.

Mathematics

2 answers:

Nina [5.8K]3 years ago

8 0

The answer is slope = 1 or m = 1
It rises 6 and goes over to the right 6 which is: 6/6
6 / 6 = 1

Hope this helps :)

pav-90 [236]3 years ago

3 0

0.8 or 0.8 over 1 because you use the formula y2-y1 over x2-x1

You might be interested in

Identify the center and the radius of the circle given by the equation (x – 4)2 + (y+13)2 = 289.

VMariaS [17]

Answer:

Center: (4,-13)

Radius: 17

Step-by-step explanation:

I guess you do not need an explanation, so I gave you only an answer.

7 0

3 years ago

54.54 54 54

Andre45 [30]

Catches me on yo yo

7 0

3 years ago

Which function will have a change in direction?

Ivahew [28]

Answer:

Quadratic

Step-by-step explanation:

A linear equation is just a line and an exponential function goes in the same direction

A quadratic us a parabola which has a turning point

3 0

3 years ago

If sinx = p and cosx = 4, work out the following forms :<br><br><br>

Kay [80]

Answer:

$\frac{p^2 - 16} {4p^2 + 16}$

Step-by-step explanation:

I will work with radians.

$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$

First, I will deal with the numerator

$\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)$

Consider the following trigonometric identities:

$\boxed{\cos\left(\frac{\pi}{2}-x \right)=\sin(x)}$

$\boxed{\sin\left(\frac{\pi}{2}-x \right)=\cos(x)}$

$\boxed{\sin(-x)=-\sin(x)}$

$\boxed{\cos(-x)=\cos(x)}$

Therefore, the numerator will be

$\sin^2(x)-\sin(x)-\cos^2(x)+\sin(x) \implies \sin^2(x)- \cos^2(x)$

Once

$\sin(x)=p$

$\cos(x)=4$

$\sin^2(x)-\cos^2(x) \implies p^2-4^2 \implies \boxed{p^2-16}$

Now let's deal with the numerator

$[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]$

Using the sum and difference identities:

$\boxed{\sin(a \pm b)=\sin(a) \cos(b) \pm \cos(a)\sin(b)}$

$\boxed{\cos(a \pm b)=\cos(a) \cos(b) \mp \sin(a)\sin(b)}$

$\sin(\pi -x) = \sin(x)$

$\sin(2\pi +x)=\sin(x)$

$\cos(2\pi-x)=\cos(x)$

Therefore,

$[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]$

$\implies [p+4] \cdot [p \cdot 4]=4p^2+16p$

The final expression will be

$\frac{p^2 - 16} {4p^2 + 16}$

8 0

3 years ago

I need help with this

Reil [10]

Answer:

see attached

Step-by-step explanation:

The domain is the horizontal extent of the graph. The graph extends to infinity in both directions horizontally (that's what the arrows mean). There are no "holes" because the open circle at x=-1 is matched by a filled circle at the same location.

The range is the vertical extent of the graph. The minimum is -3, which is included in the range. The maximum is infinity (as indicated by the up-pointing arrow).

5 0

3 years ago