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Rashid [163]
3 years ago
16

How would the graph change if the b value in the equation is decreased but remains greater than 1? Check all that apply.

Mathematics
2 answers:
Harrizon [31]3 years ago
6 0

Answer:

The graph will begin on a lower point on the y-axis.

The y-values will continue to increase as x increases.

Step-by-step explanation:

avanturin [10]3 years ago
5 0

Answer:

The graph will begin on a lower point on the y-axis.

The y-values will continue to increase as x increases

Step-by-step explanation:

just did it on edge

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PLEASE HELP ME WITH THIS QUESTION AND ILL BRAINLIEST YOU
neonofarm [45]

Answer:

c

Step-by-step explanation:

Two sum of any two side lengths of a triangle will always be greater than the length of the remaining side

a doesn't work because 8+12=20 and 20 is not greater than 20

b doesn't work because 4+5=9 and 9<10

d doesn't work because 6+6=12 and 12 is not greater than 12

c is the only option where you can take any two sides, add them, and the sum will be greater than the remaining side.

For example, for c,

6+8=14, and 14>10

8+10=18, and 18>6

10+6=16, and 16>8

7 0
3 years ago
Read 2 more answers
PLEASE HELP!!!!!! I need help on this problem
Thepotemich [5.8K]

Answer:

185 minutes.

Step-by-step explanation:

It is possible to model an equation for this, such that

0.10x + 14.80 = 0.08x + 18.50

where x represents the number of minutes.

Then, you solve for x:

0.10x + 14.80 = 0.08x + 18.50 \\ 0.02x = 3.70 \\ x =  \frac{3.70}{ 0.02}  \\ x = 185

5 0
3 years ago
If f(2) = 13 and f '(x) ≥ 3 for 2 ≤ x ≤ 4, how small can f(4) possibly be?
aniked [119]
The slope f'(x) = [f(4) - f(2)]/(4-2)≥3,
so [f(4) - 13]/2 ≥3
f(4) -13 ≥ 6
f(4)≥19, so it can be as small as 19.
8 0
3 years ago
Pls help! TIMED test!​
enyata [817]

Answer:

2

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What is the value of x and the length of the segment?
mylen [45]

Answer:

5x - 13 = x+19

5x-x=19+13

4x=32

x=32/4

x=8

Cb=5x-13

=5×8-3

=40-3

CB=37

AB=X+19

=8+19

AB=27

7 0
3 years ago
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