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GREYUIT [131]
3 years ago
10

Can someone help me pls 10 points

Mathematics
2 answers:
AnnZ [28]3 years ago
8 0

Answer:

3/5

Step-by-step explanation:

Slope = rise/run = (4-1)(1- -4) = 3/5.

EastWind [94]3 years ago
5 0
The slope is rise over run
3/5
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PLEASE HELLLPPPPPP WILL GIVE BRAINLIEST
MatroZZZ [7]
Agree tbh cause yes mhm
5 0
3 years ago
Find the midpoint between the two points (-4,4) and (-2,2)
Ymorist [56]

Answer: (-3,3) is the answer

6 0
3 years ago
I'll give brainliest if you show your work. Plz answer quickly.
Maslowich

Answer:

  C) In step 3, he should have determined which square is closest to 18.

Step-by-step explanation:

17.64 also rounds to 18. 17.64 differs from 18 by ...

  18 -17.64 = 0.36

whereas 18.49 differs from 18 by ...

  18.49 -18 = 0.49

Hence 17.64 is closer to 18 and we might expect 4.2 to be closer to √18 than is 4.3.

___

The actual root of 18 is about 4.243, so 4.2 is a better approximation.

5 0
3 years ago
Read 2 more answers
if f is a differentiable function and f(0)=-1 and f(4)-3 then which of the following must be true there exists a c in [0,4] wher
iragen [17]

Answer:

True, see proof below.

Step-by-step explanation:

Remember two theorems about continuity:

  1. If f is differentiable at the point p, then f is continuous at p. This also applies to intervals instead of points.
  2. (Bolzano) If f is continuous in an interval [a,b] and there exists x,y∈[a,b] such that f(x)<0<f(y), then there exists some c∈[a,b] such that f(c)=0.

If f is differentiable in [0,4], then f is continuous in [0,4] (by 1). Now, f(0)=-1<0 and f(4)=3>0. Thus, we have the inequality f(0)<0<f(4). By Bolzano's theorem, there exists some c∈[0,4] such that f(c)=0.

7 0
3 years ago
What two rational expressions sum to 3x+4/x^2-6x+5?
Lubov Fominskaja [6]

Answer:

\frac{3x+4}{x^2-6x+5}=\frac{-7}{4(x-1)} +\frac{19}{4(x-5)}

Step-by-step explanation:

The given rational expression is:

\frac{3x+4}{x^{2}-6x+5} = \frac{3x+4}{(x-1)(x-5)}

We can use concept of Partial Fractions to solve this problem. Let,

\frac{3x+4}{(x-1)(x-5)}=\frac{A}{x-1} +\frac{B}{x-5}

Multiplying both sides by (x - 1)(x - 5), we get:

3x+4=A(x-5)+B(x-1)

Substituting x = 5, we get:

3(5)+4=A(5-5)+B(5-1)\\\\ 15+4=0+4B\\\\ 19=4B\\\\ B=\frac{19}{4}

Substituting x = 1, we get:

3(1)+4=A(1-5)+B(1-1)\\\\ 7=-4A\\\\ A=-\frac{7}{4}

Substituting the value of A and B, back in the original equation, we get:

\frac{3x+4}{x^2-6x+5}=\frac{-7}{4(x-1)} +\frac{19}{4(x-5)}

4 0
3 years ago
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