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Leokris [45]
3 years ago
7

-1. A coffee shop spends $400 for an order of 20 cases of paper cups. How much is

Mathematics
1 answer:
tamaranim1 [39]3 years ago
7 0

Answer:

$20

Step-by-step explanation:

400/20=20

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Given that f(x)=2x+1 and g(x)=-5x+2 slice for f(g(x)) when x=3?
mezya [45]

Answer:

x = -25.

Step-by-step explanation:

If f(x) = 2x + 1, and g(x) = -5x + 2, then f(g(x)) = 2(-5x + 2) + 1 = -10x + 4 + 1 = -10x + 5.

f(g(x)) = -10x + 5, and x = 3.

-10 * 3 + 5

= -30 + 5

= -25.

Hope this helps!

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3 years ago
Pls help: √5 multiplied by √245
Luden [163]

I believe it is 35!

I hope this helps you, and don't forget to give this answer brainliest if it really helped!

5 0
2 years ago
When yeast is combined with flour and other ingredients to make pizza dough is a _____.
mihalych1998 [28]

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Chemical

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4 0
3 years ago
Read 2 more answers
Algebra 2 help!! Find the value of k
Oksana_A [137]

13  By the factor theorem, if the function is divisible by x - 3 then f(3) = 0. So we have:-

2(3)^3 + k(3)^2 + 7(3) - 3 = 0

9k = 3 - 21 - 54 = -72

k =  -72/9 =  -8 answer

14.   Similarly ,   f(-4) = 0 , so we have:-

(-4)^3 + 9(-4)^2 + k(-4) - 12 = 0

-64 + 144 - 4k - 12 = 0

4k = -64 + 144 - 12 = 68

k = 68/4

k = 17   answer

4 0
3 years ago
Verify that the function <img src="https://tex.z-dn.net/?f=g%28x%29%3D2x%5E3-3x%2B1" id="TexFormula1" title="g(x)=2x^3-3x+1" alt
ch4aika [34]

Lets check if the three conditions hold.

<u>1 : Continuity of g on the interval [0,2]</u>

First, g(x) is a continuous function on R, as the sum of a cubic function wich is continuous on R, and a linear polynomial of the form ax + b which is also continuous on R. Finally g is also continuous on the interval [0,2]

<u>2 : Differentiable on the same interval</u>

Since the cubic function and the linear polynomial one are differentiable on R, g also is differentiable and particularly on the interval [0,2]

Also we have g'(x) = 2*3*x² - 3 = 6x² - 3

<u>3 : Do we have g(0) = g(2) ?</u>

Lets compute g(0) = 2*0^3 - 3*0 + 1 = 1

And g(2) = 2*2^3 - 3*2 + 1 = 2 * 8 - 6 + 1 = 16 - 6 + 1 = 11

Since g(0) ≠ g(2), Rolle's theorem is not applicable. Thus unfortunately, we can not conclude that there exist c ∈ (0,2) such that f'(c) = 0

5 0
2 years ago
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