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iogann1982 [59]
4 years ago
9

Suppose you invest $500 in a savings account that pays 3.5% annual interest. When will the account contain at least $650?

Mathematics
1 answer:
Anvisha [2.4K]4 years ago
4 0
Start with 650-500, then multiply by .035. So the entire equation would be 650-500= 150x.035= 5.25
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a school teacher can grade 3 tests in 18 minutes. If class starts in 3 hours and there are 27 tests to grade, will they be done
lara31 [8.8K]

Answer:

yes.

honestly, I didn't really do all the math.

3 in 18 minutes, a little better than 3 in 20 minutes. that translates to 9 in an hour, with 6 minutes left over.

3 hours of that (3×9) is the 27. with 6 minutes per hour left over, for a total of 18 minutes.

the teacher has time to grade the tests, get a cup of coffee, and take a few minutes to wonder why she didn't go into an easier line if work.

8 0
3 years ago
Cooper is measuring ingredients for his special dessert. He needs a total of 70.4 mL of milk and melted butter. The recipe calls
AysviL [449]
Milk (m) , butter (b)

m + b = 70.4
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6 0
4 years ago
Solve the indicated variable for this literal equation -10 = xy + z for x
Fiesta28 [93]
X=y+z+10 you have to move the x by itself of the equal sign
6 0
3 years ago
Read 2 more answers
Mr. Anderson is ordering pizzas for a class pizza party. Pizza Place has a special where he can buy 3 large pizzas for $18.75. A
lukranit [14]

Answer:

They would save 9 dollars with Mario's pizza

Step-by-step explanation:

22*3 is 66

18.75*4 is 75

75-66=9

8 0
3 years ago
A certain car depreciates such that its value at the end of each year is p % less than its value at the end of the previous year
icang [17]

Answer:

b(b/a)^2

Step-by-step explanation:

Given that the value of the car depreciates such that its value at the end of each year is p % less than its value at the end of the previous year and that car was worth a dollars on December 31, 2010 and was worth b dollars on December 31, 2011, then

b = a - (p% × a) = a(1-p%)

b/a = 1 - p%

p% = 1 - b/a = (a-b)/a

Let the worth of the car on December 31, 2012 be c

then

c = b - (b × p%) = b(1-p%)

Let the worth of the car on December 31, 2013 be d

then

d = c - (c × p%)

d = c(1-p%)

d = b(1-p%)(1-p%)

d = b(1-p%)^2

d = b(1- (a-b)/a)^2

d = b((a-a+b)/a)^2

d = b(b/a)^2 = b^3/a^2

The car's worth on December 31, 2013 =  b(b/a)^2 = b^3/a^2

4 0
4 years ago
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