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choli [55]
3 years ago
15

Anthony will be competing in a body building competition in 12 months. he initially weighs 160 pounds and by the end of 12 month

s weighs 232 pounds. What is the approximate rate of change in Anthony's body weight?
Mathematics
1 answer:
padilas [110]3 years ago
4 0
He starts out with the weight of 160, at the end of those 12 months he weighs 232. if you subtract 160 and 232 the approximate rate of change should be 72
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Answer:

m =-6

Step-by-step explanation:

To solve for the unknown variable, m, we have to get m by itself. Perform the opposite of what is being done to the equation.

-8m-9=39

9 is being subtracted from-8m, so add 9 to both sides. Addition is the opposite of subtraction.

-8m-9+9=39+9

-8m=39+9

-8m=48

-8 and m are being multiplied. Divide both sides by -8, because division is the opposite of multiplication

-8m/-8=48/-8

m=48/-8

m= -6

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It depends on how much sun the place receives, how much it is shaded. temperature in the area. Do you have solar panels, electric heat, gas? All these factors determine utility cost.

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atroni [7]

Answer:

x>9

Step-by-step explanation:

Tamika already has $55, which will stay the same independent of how much time passes. Since she deposits $5 per week, we can assign weeks a variable x. The amount of money in her account after x weeks is equal to 55+5x. If she must save over $100, we can create an inequality to solve for how many weeks must pass: 55+5x>100.

The question also asks to solve the inequality:

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  3. 5x>45
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Step-by-step explanation:

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How many gallons of a 80% antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get a mixture that is 70% antif
konstantin123 [22]

440 gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get a mixture that is 70% antifreeze

<em><u>Solution:</u></em>

Let "x" be the gallons of 80 % antifreeze added

Therefore, "x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze

Final mixture is x + 80

Therefore, we can frame a equation as:

"x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get (x + 80) gallons of 70 % antifreeze

Thus, we get,

x gallons of 80 % + 80 gallons of 15 % = (x + 80) gallons of 70 %

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Thus 440 gallons of 80 % antifreeze solution must be mixed

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