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USPshnik [31]
3 years ago
11

Rewrite the following equation as a function of x. 680x + 10y -1000 = 0

Mathematics
1 answer:
hodyreva [135]3 years ago
5 0

Answer:

y= -68 x + 100

Step-by-step explanation:

subject y as in the form of y= mx+ c

we get

10 y= -680 x + 1000

now we can divide every term by 10 as to obtain least values

Therefore :y= -68 x+ 100

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123

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What is the unknown fraction?
daser333 [38]

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A. 31/100

Step-by-step explanation:

Change 3/10 to hundreds:

3/10 × 10/10 = 30/100

The denominator (the bottom number) must be 100.

61 - 30 is 31.

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X + 1/4x = 15. What is the solution to the equation?
Sauron [17]

Answer:

Step-by-step explanation:

hello :

X + 1/4x = 15

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7 0
3 years ago
In each of Problems 5 through 10, verify that each given function is a solution of the differential equation.
WARRIOR [948]

Answer:

For First Solution: y_1(t)=e^t

y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:y_2(t)=cosht

y_2(t)=cosht  is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: y_1(t)=e^t

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_1(t)=e^t

First order derivative:

y'_1(t)=e^t

2nd order Derivative:

y''_1(t)=e^t

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:

y_2(t)=cosht

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_2(t)=cosht

First order derivative:

y'_2(t)=sinht

2nd order Derivative:

y''_2(t)=cosht

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence y_2(t)=cosht  is the solution of equation y''-y=0.

3 0
3 years ago
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