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Westkost [7]
4 years ago
12

Fill in the blanks to translate the following English phrase into a variable expression

Mathematics
1 answer:
igomit [66]4 years ago
8 0
8z+3

eight times the number z = 8z

add 3 to make it three more.
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A business owes $8000 on a loan. Every month, the business pays 1/2 of the amount remaining on the loan.
sveticcg [70]
Here is the answer to the given problem above. Given that the business owes $8000 on a loan and every month, it pays 1212. So for the first month, the remaining is 6,788. On the second month is 5,576, and on the third month, it is 4,364. In the fourth month, the remaining would be 3,152, and in the fifth month is 1,940. Therefore, on the sixth month, if the business pays 1,212 still, the remaining would be $728 only. Hope this answer helps.
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If 7m = 35, then 7m + 4 = 35 + 4 WHAT IS THE PROPERTY?
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Answer:

Addition Property

because it has both multiplication and addition but the addition is more

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Which of these is the algebraic expression for "six more than the product of seven and some number?"
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7x+6 is the algebraic expression represented
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For z=-3-5i, which graph shows z and the product z•i?
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Answer: it’s A or the 1st option

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
H=54+1/2at^2 solve for a<br> Im lost on if I should square the 1/2at first?
Verdich [7]
<h3>Answer:  a = (2h-108)/(t^2) </h3>

Other answer formats are possible.

===========================================================

Explanation:

Let's say we want to solve the equation h = 54 + 10a for the variable 'a'. This is relevant to the problem at hand don't worry.

What we'd do is two things

  1. Subtract 54 from both sides
  2. Divide by 10

So the steps look like this

h = 54 + 10a

h-54 = 10a

(h-54)/10 = a

a = (h-54)/10

Now let's say we wanted to solve this equation: h = 54 + ca, where c is a constant and can stand in place of any number. The steps would be pretty much identical. Instead of dividing both sides by 10, we divide both sides by c. We can do this as long as c isn't zero of course.

That means the equation h = 54+ca solves to a = (h-54)/c

---------------------------

Why did I do those examples? To build up to the equation your teacher gave you.

Let's replace c with 0.5t^2

This is because 1/2 = 0.5

The 1/2at^2 is the same as (1/2t^2)a

So instead of a = (h-54)/c, we'd get a = (h-54)/(0.5t^2)

As a last optional and encouraged step, we can multiply every term by 2 to get rid of that decimal value.

Doing so makes a = (h-54)/(0.5t^2) turn into a = (2h-108)/(t^2)

---------------------------

Here's another way to solve:

h = 54 + (1/2)at^2

h-54 = (1/2)at^2

2(h-54) = at^2

2h-108 = at^2

(2h-108)/(t^2) = a

a = (2h-108)/(t^2)

In the second step, I subtracted 54 from both sides. Immediately afterward, I multiplied both sides by 2 so I could clear out the fraction. Then I divided both sides by t^2 to fully isolated the variable 'a'.

6 0
3 years ago
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