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Gnesinka [82]
3 years ago
12

IDENTIFY EACH COEFFICIENT: 3m

Mathematics
1 answer:
Rainbow [258]3 years ago
4 0

Answer:

theirs nothing to identify tho

Step-by-step explanation:

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Can anyone help me please and explain this: Write "60 pounds for $6.60" as a unit rate per ounce to the nearest tenth. 1 pound=
Pepsi [2]
<span>1 pound = 16 ounces
so 60 pounds = 60 x 16 = 960 ounces.
$6.60/960 =0.006875 per ounce.
0.006875 x 16 ounces = 0.11 (11 cents per pound)

<span>3 pounds x 0.11= 33 cents 

answer is B</span></span>
6 0
3 years ago
A repair company's charge for repairing a certain type of copy machine fits the model y = 47.38 + 0.617x where y is the number o
11Alexandr11 [23.1K]

Answer:c is the correct answer

Step-by-step explanation:

A repair company's charge for repairing a certain type of copy machine fits the model y = 47.38 + 0.617x

where y is the number of dollars charged and

x is the number of minutes the repair person is on the job.

Therefore, to determine the number of minutes that it would take for the cost of repair to reach $130, we would substitute y =130 into the given model. It becomes

130 = 47.38 + 0.617x

0.617x = 130 - 47.38 = 82.62

x = 82.62/0.617 = 134 minutes

4 0
3 years ago
Are these right?<br>9. 6<br>10. 7<br>11. 8<br>12. 8<br>13. 10<br>14. 8<br>15. 9<br>16. 10
andrey2020 [161]
Yes, again, they are correct.
4 0
3 years ago
Your starter to find the slope of the line
sergejj [24]
Slope is 3/1 (3 simplified)
4 0
3 years ago
Prove cosh 3x = 4 cosh^3 x - 3 cosh x.
Snezhnost [94]
Prove we are to prove  4(coshx)^3 - 3(coshx) we are asked to prove 4(coshx)^3 - 3(coshx) to be equal to cosh 3x
= 4(e^x+e^(-x))^3/8 - 3(e^x+e^(-x))/2 = e^3x /2 +3e^x /2 + 3e^(-x) /2 + e^(-3x) /2 - 3(e^x+e^(-x))/2 = e^(3x) /2 + e^(-3x) /2 = cosh(3x) = LHS Since y = cosh x satisfies the equation if we replace the "2" with cosh3x, we require cosh 3x = 2 for the solution to work. 
i.e. e^(3x)/2 + e^(-3x)/2 = 2 
Setting e^(3x) = u, we have u^2 + 1 - 4u = 0 
u = (4 + sqrt(12)) / 2 = 2 + sqrt(3), so x = ln((2+sqrt(3))/2) /3, Or u = (4 - sqrt(12)) / 2 = 2 - sqrt(3), so x = ln((2-sqrt(3))/2) /3, 
Therefore, y = cosh x = e^(ln((2+sqrt(3))/2) /3) /2 + e^(-ln((2+sqrt(3))/2) /3) /2 = (2+sqrt(3))^(1/3) / 2 + (-2-sqrt(3))^(1/3) to be equ
= 4(e^x+e^(-x))^3/8 - 3(e^x+e^(-x))/2 
= e^3x /2 +3e^x /2 + 3e^(-x) /2 + e^(-3x) /2 - 3(e^x+e^(-x))/2 
= e^(3x) /2 + e^(-3x) /2 
= cosh(3x) 
= LHS 

<span>Therefore, because y = cosh x satisfies the equation IF we replace the "2" with cosh3x, we require cosh 3x = 2 for the solution to work. </span>

i.e. e^(3x)/2 + e^(-3x)/2 = 2 

Setting e^(3x) = u, we have u^2 + 1 - 4u = 0 

u = (4 + sqrt(12)) / 2 = 2 + sqrt(3), so x = ln((2+sqrt(3))/2) /3, 
Or u = (4 - sqrt(12)) / 2 = 2 - sqrt(3), so x = ln((2-sqrt(3))/2) /3, 

Therefore, y = cosh x = e^(ln((2+sqrt(3))/2) /3) /2 + e^(-ln((2+sqrt(3))/2) /3) /2 
= (2+sqrt(3))^(1/3) / 2 + (-2-sqrt(3))^(1/3)
3 0
4 years ago
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