All categories

3 years ago

What is the slope of the line through (-5,-2) and (-1,-10)?

Mathematics

2 answers:

joja [24]3 years ago

8 0

Answer:

m=-2

Step-by-step explanation:

plug in the points in the formula y2-y1/x2-x1

I suggest getting m a t h w a y its a website and app thats useful.. and good.. and I like it

kicyunya [14]3 years ago

4 0

answer

-2

explanation

the slope can be found through m = (y2 - y1) / (x2 - x1)

with the points (-5, -2) and (-1, -10)

x1 = -5

x2 = -1

y1 = -2

y2 = -10

m = (-10 - -2) / (-1 - -5)

m = -8 / 4

m = -2

You might be interested in

**WILL GIVE BRAINLIEST** MUST SHOW WORK!

Degger [83]

I THINK IT C

I REALLY TRIED

HOPE I HELPED

8 0

3 years ago

Read 2 more answers

Math help on this question!

NemiM [27]

Answer:

none

Step-by-step explanation:

110 + 60 = 170

but 170 is obviously not 180 sooo none will be 60 degrees

6 0

3 years ago

Ian bought a cup of coffee. At first, he sipped slowly. As it cooled down, he drank more quickly. The last bit was cold, and he

Colt1911 [192]

Answer:

ian just wasted the last bit of coffee.

Step-by-step explanation:

4 0

3 years ago

(Geometry)

lianna [129]

Answer:

Is it clockwise rotation or counterclock wise?

If it’s clockwise rotation then C is the answer but if it’s counter lock wise rotation then the answer is A.

Hope this helps!

3 0

3 years ago

Let {v1,v2,v3} be a set of vectors in Rn . If u is Span {v1,v2,v3}, show that 3u is in Span {v1,v2,v3}.(Hint: since you are told

ANTONII [103]

Answer:

See proof below

Step-by-step explanation:

We will use the hint. The statement of the hint holds true, as the linear span of a set of vectors T is equal to the set of linear combinations of vectors in T.

Denote the linear span of vectors with the curly brackets < >, that is, $span\{v_1,v_2,v_3\}:=$

Let $u\in$ , then u is a linear combination of v1,v2,v3, that is, there exist scalars $c_1,c_2,c_3\in \mathbb{R}$ such that $u=c_1v_1+c_2v_2+c_3v_3$ . Multiply by 3 in both sides to get $3u=3c_1v_1+3c_2v_2+3c_3v_3=d_1v_1+d_2v_2+d_3v_3$ , with $d_i=3c_i,i=1,2,3$

Since $c_1,c_2,c_3\in \mathbb{R}$ , $d_1,d_2,d_3\in \mathbb{R}$ as real numbers are closed under multiplication. Therefore 3u is a linear combination of the vectors $v_1,v_2,v_3$ , that is, $3u\in$

8 0

3 years ago