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Sauron [17]
3 years ago
14

Solve for x A. 9 B. 10 C. 42 D. 12

Mathematics
2 answers:
Eva8 [605]3 years ago
8 0

Answer:

Given:-  

m∠DEF= (4x-8) °

m DE= 80°

Using a property :- m ∠DEF= 1/2 (m DE)

(4x-8) ° = 1/2 (80)

(4x-8)° =(40) °

4x-8=40

Add 8 to both sides:-

4x=48

Divide both sides by 4:-

\frac{4x}{4}=\frac{48}{4}

x=12

<u>OAmalOHopeO</u>

Cloud [144]3 years ago
4 0

Answer:

Option D, 12

Step-by-step explanation:

4x-8=80/2

or, 4x-8=40

or, 4x=48

or, x=12

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Kristen invests $ 5745 in a bank. The bank 6.5% interest compounded monthly. How long must she leave the money in the bank for i
podryga [215]

Answer:

1. 10.7 years

2. 17.0 years

3. 2nd option

Step-by-step explanation:

Use formula for compounded interest

A=P\cdot \left(1+\dfrac{r}{n}\right)^{nt},

where

A is final value, P is initial value, r is interest rate (as decimal), n is number of periods and t is number of years.

In your case,

1. P=$5745, A=2P=$11490, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then

11490=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},\\ \\2=(1.0054)^{12t},\\ \\12t=\log_{1.0054}2,\\ \\t=\dfrac{1}{12}\log_ {1.0054}2\approx 10.7\ years.

2. P=$5745, A=3P=$17235, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then


17235=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},

3=(1.0054)^{12t},

12t=\log_{1.0054}3,

t=\dfrac{1}{12}\log_{1.0054}3\approx 17.0\ years.

3. <u>1 choice:</u> P=$5745, n=12 (compounded monthly), r=0.065 (6.5%), t=5 years and A is unknown. Then

A=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12\cdot 5},\\ \\A=5745\cdot (1.0054)^{60}\approx \$7936.39.

<u>2 choice:</u> P=$5745, n=4 (compounded monthly), r=0.0675 (6.75%), t=5 years and A is unknown. Then

A=5745\cdot \left(1+\dfrac{0.0675}{4}\right)^{4\cdot 5},\\ \\A=5745\cdot (1.0169)^{20}\approx \$8032.58.

The best will be 2nd option, because $8032.58>$7936.39

5 0
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Answer:

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Fittoniya [83]

Answer:

2592 cm

Step-by-step explanation:

square area = x²

rectangle area = bh

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x² = bh

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b and h are in the problem as 64 and 81

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<em>i</em><em> </em><em>hope</em><em> </em><em>it</em><em> </em><em>helped</em><em>.</em><em>.</em><em>.</em>

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