All categories

3 years ago

Find the equation of a line that passes through the points (-1,-5) and (-3,-4). Leave your answer in the form y=mx+c

Mathematics

1 answer:

MaRussiya [10]3 years ago

8 0

Answer:

y = -1/2x - 11/2

Step-by-step explanation:

y2 - y1 / x2 - x1

-4 - (-5) / -3 - (-1)

1/ -2

= -1/2

y = -1/2x + b

-5 = -1/2(-1) + b

-5 = 1/2 + b

-11/2 = b

You might be interested in

What is 5c - 6 = 4 -3c, im blanking on what im supposed to do

Firlakuza [10]

Step-by-step explanation:

im pretty sure ur supposed to find so the answer woild be c=0.8

7 1

4 years ago

Read 2 more answers

Stephen is 5'7". Him and his class mates want to know how many Stephens it would take to get to The Sun (4.8 billion feet from t

Sergio [31]

5'7" is Stephen's height

4.8 billion is the distance to the sun.

4.8 billion ft / 5'7" = 859,701,493 ft

6 0

4 years ago

Which point lies on the line shown in the graph?

CaHeK987 [17]

(7,0) because that’s where it is crossing the x-intercept

8 0

3 years ago

Read 2 more answers

Simplify 4(6 - 2x) - 3(3 - 4x)<br><br> 4x + 15<br> 15 - 20x<br> 15 - 6x

Leto [7]

Answer:

4x+15 Is the answer

Step-by-step explanation:

4(6−2x)−3(3−4x)

Distribute:

=(4)(6)+(4)(−2x)+(−3)(3)+(−3)(−4x)

=24+−8x+−9+12x

Combine Like Terms:

=24+−8x+−9+12x

=(−8x+12x)+(24+−9)

=4x+15

8 0

3 years ago

Read 2 more answers

Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

vodka [1.7K]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

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

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

7 0

3 years ago