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Anastasy [175]
3 years ago
15

The total of two investments are $5,000. The average annual interest rate for one investment is 10% and 14% for the other. The t

otal annual interest is $620. How much is invested at each rate.
Mathematics
1 answer:
olga_2 [115]3 years ago
8 0

Answer:

Step-by-step explanation:

How  the heck imma suppose to no

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6x+14x+5=5(4x+1) what is x in this equation
andrew-mc [135]
Infinite many solutions
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Ella grew 3 over 4 of an inch last year. Ben grew 2 over 3 of an inch last year. How much more did Ella grow than Ben last year?
IrinaK [193]

Answer:

Ella grew \frac{1}{12} of an inch more

Step-by-step explanation:

For this problem, you want to subtract Ben's growth from Ella's.

This expression would be \frac{3}{4}-\frac{2}{3}.

To work these out, they must have the same denominator (which should be the lowest common multiple or LCM).

In this case, the LCM is 12, so you want to multiply each side so the denominator is 12. This means the first fraction should be multiplied by \frac{3}{3} and the second by \frac{4}{4}.

This makes the expression \frac{9}{12}-\frac{8}{12}, which equals \frac{1}{12}.

**This content involves adding and subtracting fractions, which you may want to revise. I'm always happy to help!

7 0
3 years ago
What is 15.75 rounded to the whole number​
victus00 [196]

Answer:

C. 16

Step-by-step explanation:

4 0
3 years ago
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How to work it out logically? Question a !!!
Arturiano [62]

Answer:

Fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.

Step-by-step explanation:

Given the figure with dimensions. we have to find the area of given figure.

Area of figure=ar(1)+ar(2)+ar(3)

Area of region 1 = ar(ANGI)+ar(AIB)

                          =L\times B+\frac{1}{2}\times base\times height\\\\=[1500\times (5000-2000-1500)]+\frac{1}{2}\times (3000-1500)\times (5000-2000-1500)\\\\=3375000m^2=337.5ha

Area of region 2 = ar(DHBC)

                       =2000\times1500\\\\=3000000m^2=300ha

Area of region 3 = ar(GFEH)

                             (2000+1500)\times 1000\\\\=3500000m^2=350ha

Hence, Area of figure=ar(1)+ar(2)+ar(3)=337.5ha+300ha+350ha

                                                 =987.5 ha

Now, we have to do straight-line fencing such that area become half and cost of fencing is minimum.

Let the fencing be done through x m downward from B which divides the two into equal area.

⇒ Area of upper part above fencing=Area of lower part below fencing

⇒ar(ANGB)+ar(GKLB)=ar(KLCM)+ar(MDCF)\\\\337500+3000x=(3500-x)\times 1000+2000(1500-x)\\\\3375000+3000x=3500000-1000x+3000000-2000x\\\\6000x=315000\\\\x=\frac{315000}{6000}=520.8m

Hence, fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.

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3 years ago
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I dont understand this problem. Could someone explain it to me?
IRINA_888 [86]

Answer:

additive property

Step-by-step explanation:

it is because when you ADD the angles it is EQUAL to the answer

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3 years ago
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