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qwelly [4]
3 years ago
13

Given the function h(x) = x2 – 7x + 8, determine the average rate of change of

Mathematics
2 answers:
Kryger [21]3 years ago
8 0

Answer:

Step-by-step explanation:

-2

True [87]3 years ago
6 0

Answer:

4

Step-by-step explanation:

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To find the midpoint of a segment on the coordinate plane, you find the
Vaselesa [24]

Answer:

A.True

Step-by-step explanation:

To find the midpoint of a line  segment, we find the averages/mean of the endpoints of the line segment.

Midpoint Formula:

M(x,y)=(x₁+x₂/2 , y₁+y₂/2)

6 0
3 years ago
Read 2 more answers
Nia's bag weighs 5-sixths of the maximum weight allowed for bags on an airplane. The maximum weight allowed is 45 pounds. What i
scZoUnD [109]

Answer:

54

Step-by-step explanation:

copy  dot flip 45/1 divided by 5/6

first copy dot flip so 45/1 then multiply and flip 5/6 so it looks like this

45/1 x 6/5= 270/5 then you need to see how many time 5 goes in 270 which is 54 and thats you anwser hope this helps sorry if im wrong

4 0
3 years ago
QUIZ:
Reika [66]

Answer:

idk

Step-by-step explanation:

i have no clue

6 0
3 years ago
. Find the inverse of the function below on the given interval and write it in the form yequalsf Superscript negative 1 Baseline
Elena L [17]

Answer:

The inverse of the function is f^{-1}(x)=\frac{x-5}{3}.

Step-by-step explanation:

The function provided is:

f (x)=3x+5

Let f(x)=y.

Then the value of <em>x</em> is:

y=3x+5\\\\3x=y-5\\\\x=\frac{y-5}{3}

For the inverse of the function, x\rightarrow y.

⇒ f^{-1}(x)=\frac{x-5}{3}

Compute the value of f[f^{-1}(x)] as follows:

f[f^{-1}(x)]=f[\frac{x-5}{3}]

               =3[\frac{x-5}{3}]+5\\\\=x-5+5\\\\=x

Hence proved that f[f^{-1}(x)]=x.

Compute the value of f^{-1}[f(x)] as follows:

f^{-1}[f(x)]=f^{-1}[3x+5]

               =\frac{(3x+5)-5}{3}\\\\=\frac{3x+5-5}{3}\\\\=x

Hence proved that f^{-1}[f(x)]=x.

8 0
3 years ago
25 points
VMariaS [17]

Answer:

5 and 7

Step-by-step explanation:

because the problem said least two girls meaning substract

6 0
2 years ago
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