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ICE Princess25 [194]

3 years ago

11

What do you add to 2 7/8 to make 5

1 answer:

qwelly [4]3 years ago

5 0

Answer:

2 1/8

Step-by-step explanation:

7/8 is the same as 0.875 and therefore you need 0.125 also known as 1/8 to make it a whole number. When you add it to the already existing whole 2 you get three. Subtract three from five to make two which is what you need to add on top to finally get 5.

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-6(y + 15) = -3y + 6 what is the value of y

Serga [27]

Let's begin by listing out the information given to us:

$-6\mleft(y+15\mright)=-3y+6$

We are to find the value of y

$undefined$

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Citrus2011 [14]

If I’m correct it should be X=3

6 0

3 years ago

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Why are the solutions on the interval 0 < x < 360 <br> Solve the equation cos (x/2) = cos x+ 1

garri49 [273]

Answer:

lesss go

Step-by-step explanation:

6 0

3 years ago

For y = 3x - 2x2 +5x3, show how to find the value for y, when x = 3. Show your work

Lelechka [254]

Answer:

y= 126

Step-by-step explanation:

The given equation is: y= 3x-2x^2+5x^3

Substituting the value x=3 in the above equation,we get

y= 3(3)-2(3)^2+5(3)^3

⇒y= 9-2(9)+5(27)

⇒y= 9-18+135

⇒y= 126

5 0

3 years ago

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A substance with a half life is decaying exponentially. If there are initially 12 grams of the substance and after 70 minutes th

77julia77 [94]

Answer: 233 min

Step-by-step explanation:

This problem can be solved by the following equation:

$A=A_{o} e^{-kt}$ (1)

Where:

$A=7 g$ is the quantity left after time $t$

$A_{o}=12 g$ is the initial quantity

$t=70 min$ is the time elapsed

$k$ is the constant of decay for the material

So, firstly we need to find the value of $k$ from (1) in order to move to the next part of the problem:

$\frac{A}{A_{o}}=e^{-kt}$ (2)

Applying natural logarithm on both sides of the equation:

$ln(\frac{A}{A_{o}})=ln(e^{-kt})$ (3)

$ln(\frac{A}{A_{o}})=-kt$ (4)

$k=-\frac{ln(\frac{A}{A_{o}})}{t}$ (5)

$k=-\frac{ln(\frac{7 g}{12 g})}{70 min}$ (6)

$k=0.00769995 min^{-1}$ (7) Now that we have the value of $k$ we can solve the other part of this problem: Find the time $t$ for $A=2 g$ .

In this case we need to isolate $t$ from (1):

$t=-\frac{ln(\frac{A}{A_{o}})}{k}$ (8)

$t=-\frac{ln(\frac{2 g}{12 g})}{0.00769995 min^{-1}}$ (9)

Finally:

$t=232.697 min \approx 233 min$

5 0

3 years ago