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Savatey [412]
2 years ago
7

Find the great common factor of 33, 63, and 90.

Mathematics
1 answer:
Alisiya [41]2 years ago
3 0

Answer:

GCF is 3

Step-by-step explanation:

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X²=169 extracting square root​
ryzh [129]

Answer:

x = ± 13

Step-by-step explanation:

Given

x² = 169 ( take the square root of both sides )

x = ± \sqrt{169} = ± 13

Since 13 × 13 = 169 and - 13× - 13 = 169

7 0
3 years ago
At the beginning of each of her four years in college, Miranda took out a new Stafford loan. Each loan had a principal of $5,500
kaheart [24]

Answer:

D. $31,337.27

Step-by-step explanation:

We have that the initial amount of the loan is $5500.

Miranda took the loan for 4 years. So, the total present value is $5500×4 = $22,000.

The rate of interest on the loan is 7.5% i.e. 0.075 and it was for the duration of 10 years.

Also, it is given that the loan was compounded annually.

We have the formula as,

P=\frac{\frac{r}{n}\times PV}{1-(1+\frac{r}{n})^{-t\times n}}

i.e. PV=\frac{P\times [1-(1+\frac{r}{n})^{-t\times n}]}{\frac{r}{n}}

Substituting the values, we get,

i.e. PV=\frac{P\times [1-(1+\frac{0.075}{12})^{-10\times 12}]}{\frac{0.075}{12}}

i.e. 22000=\frac{P\times [1-(1+0.00625)^{-120}]}{0.00625}

i.e. 22000=\frac{P\times [1-(1.00625)^{-120}]}{0.00625}

i.e. 22000=\frac{P\times [1-0.4735]}{0.00625}

i.e. 22000=\frac{P\times 0.5265}{0.00625}

i.e. P=\frac{22000\times 0.00625}{0.5265}

i.e. P=\frac{137.5}{0.5265}

i.e. P=261.16

Thus, the total lifetime cost to pay of the loans compounded annually  = 261.16 × 120 = $31,339.2

Hence, the total cost close to the answer is $31,337.27

7 0
3 years ago
Read 2 more answers
Brainliest to correct answer pls help
Vladimir79 [104]

Answer:

.

Step-by-step explanation:

4 0
2 years ago
How may cups are In a gallon
mars1129 [50]
16US cups. I think that’s the right answer
6 0
2 years ago
Read 2 more answers
I need help with finding the value of x
Lelu [443]

Answer:

x= 10

Step-by-step explanation:

(8x+5)+(3x+25)+(4x)=180

15x+30=180

15x=150

x=10

<h2><em><u>xoxo, </u></em></h2><h2><em><u>your highness...</u></em></h2>
8 0
2 years ago
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