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2 years ago

What is the answer for number 12

Mathematics

2 answers:

Jet001 [13]2 years ago

5 0

I hope this helps you

3.4/7.4+2.7/4.7

12/28+7/28

19/28

GalinKa [24]2 years ago

4 0

(3*4)+(1*7)/7*4 = 12+7/28 = 19/28

your answer is 19/28

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3 years ago

Eric took out an 80/20 mortgage to buy a house costing $175,000. The first (80%) mortgage has an interest rate of 4.75%, and the

Serhud [2]

Answer:

The total monthly mortgage payment for the house is $975.63

Step-by-step explanation:

The principle amount is $175000

80% of 175000 is = $0.8\times175000$ = $140000

20% of 175000 is = $0.2\times175000$ = $35000

Emi formula is :

$\frac{p\times r\times(1+r)^{n} }{(1+r)^{n}-1 }$

For 1st part:

p = 140000

r = 4.75/12/100=0.00395

n = $30*12=360$

Putting values in formula we get

$\frac{140000\times0.00395\times(1.00395)^{360} }{(1.00395)^{360}-1 }$

= $729.508

For 2nd part:

p = 35000

r = 7.525/12/100=0.00627

n = $30*12=360$

Putting values in formula we get

$\frac{35000\times0.00627\times(1.00627)^{360} }{(1.00627)^{360}-1 }$

= $245.301

Adding both the monthly payments:

$729.508+245.301=974.809$ dollars

This is closest to option A.

So, option A is the answer.

And for 30 years the mortgage payment will be =

$975.63\times12\times30=351226.80$ dollars

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3 years ago

Determine the greatest common factor of 15xy^7z and 30x^2y^7z

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Answer:

$15x {y}^{7} z$

Step-by-step explanation:

<h3>Factorize the Expressions</h3>

$15x {y}^{2} z = 3 \times 5 \times x \times y \times y \times y \times y \times y \times y \times y \times z$

$30 {x}^{2} {y}^{7} z = 2 \times 3 \times 5 \times x \times x \times y \times y \times y \times y \times y \times y \times y \times z$

<h3>Factor out the Common Factors</h3>

$gcf = 3 \times 5 \times x \times y \times y \times y \times y \times y \times y \times y \times z$

<h3>Multiply the Common Factors</h3>

$3 \times 5 \times x \times y \times y \times y \times y \times y \times y \times y \times z = 15x {y}^{7} z$

The Factored expressions are

$15x {y}^{7} z$

And

$15x {y}^{7} (2x)$

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3 years ago

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