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Find the savings plan balance after 17 months with an APR of 2% and monthly payments of $551

AB : BD  =  5 : 7   =   10   :  14

AC : CD  =   13 * 11  

This  implies that     A  to  D  has  24  equal parts

AB =  10  of these

AC =  13  of  these

So   BC  = 3

So

AB =  10

BC  = 3

CD =  11

So  AB  :  BC  :  CD   =     10  :  3 :  11

14280

Thank you so much!!!

Calculate the amount of interest you'll have at the end of the indicated period. 

You invest $14,000 in an account that pays simple interest of 2% for 1 year.

Hmm lets see we just find the area of a circle that is smaller then 200. Which is 153.86. then you put 153.86/200 which is about 77%.

Using  the  Law of Cosines  we  have  that

BC^2  =  AB^2  + AC^2  - 2(AB* AC) cos  ( BAC)

9^2  = 5^2 + 8^2  - 2(5 * 8)  cos (BAC)

81 = 25  + 64  - 80 cos (BAC)

cos ( BAC)  =  [ 81  - 25 - 64 ]  / -80    =    -8  / -80    =   1/10

And applying it once  more we  have  that

DE^2  = AE^2  + AD^2  - 2(AE * AD) cos (BAC)

DE^2  = 10^2  + 16^2  - 2 ( 10 * 16)  (1/10)

DE^2 = 100 + 256  - 32

DE^2  =  356  -  32

DE^2 = 324

DE  = 18

You can find that out moo.

Hmm i think what you mean is that $280 for food and $1200 a month for health insurance? And the annual car insurane is each year. If it is like that then its just 280+1200= 1480+41.68ish which equals to 1521.68ish.

Look at this article! It will give u a hint!! https://web2.0calc.com/questions/point-a-b-c-and-d-lies-on-a-straight-line

280 per month on food, 200 per month on health insurance, and 41 2/3 for car insurance. I think?

521 2/3 for all.

THAT HELPS TOO THX

2  * 2pi * r   =  pi * r^2       divide  out  pi

4 r   =   r^2        divide  by  r  on both sides

4  =  r

how big is the circle of the sprinkler?

If that is so, the 4r\(\pi\) is equal ro r^2\(\pi\) divide both sides by \(r\pi\) to get r=4

527 with a 10 on the bottom. idk what that means but i think u may know. Thats the missing number. thanks!

There are 2 bars of gold-silver alloy; one piece has 2 parts of gold to 3 parts of silver and another has 3 parts of gold to 7 parts of silver. If both bars are melted into 8 kg bar with the final gold to silver ratio of 5:11. What was the weight of the first bar?

Here's  the way I might  go about this

Call the  weight of the first bar  = A

Then the  weight of the  second  bar  =  8 - A

If  the ratio  of gold to silver in the  8 kg  bar  = 5 : 11,  then  the   weight of the  gold must  be 

(5/16) * 8  =  2.5 kg

And  the  weight  of silver  in 8kg bar   =     5.5 kg

And if  the  parts  of  gold to  silver in the  first  bar = 2 : 3,  then  gold must  be (2/5)  of the  weight of the  first  bar

And if the parts of  gold  to  silver  in the  second  bar  =  3 : 7, then gold  must  be (3/10) of the weight of the  second bar 

So  we have  this equation

(2/5) (A)  + (3/10)  (8 - A)   =    2.5      

(4/10)A  + 24/10  - (3/10)A  =  2.5

(1/10A  +  2.4  =  2.5

(1/10)A  =  1/10

A =  1  kg  =    weight of  1st  bar

The length of one side is 45, so the length of the others are 255. The sum of the ratios add up to 15, so divide 255/15 to get 17. Multiply the number (5) one to get the largest side to be 85 units, and multiply 17 times (1) to get the smallest is 17 units. 85-17 is 68!

Thank you so much!