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Dmitrij [34]
3 years ago
6

What is the answer to this 3(k−6)=15

Mathematics
1 answer:
Alchen [17]3 years ago
5 0

Answer:

11

Step-by-step explanation:

First distribute the 3 to both the numbers inside of the parentheses:

3(k)-3(6)=15

3k-18=15

Add 18 to both sides

3k=33

Divide 3 from both sides to isolate k

k=11

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30 is what percent of 80
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30/80 = 0.375(100) = 37.5%

30 is 37.5% of 80
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If a cube has a volume of 13,824 cubic meters what is the edge length
ratelena [41]
24 cubic meters because 24*24*24=13,824.
6 0
3 years ago
what is the midpoint of the segment joining the points (4,-2)and (-8,6) please show in steps how you got it
sashaice [31]
The x-coordinate of the midpoint will be the average of the two given x-coordinates, and the y-coordinate will be the average of the two given y-coordinates.
This can be written as
midpoint = ( \frac{x1+x2}{2} , \frac{y1+y2}{2}),
where the given points are (x1, y1) and x2, y2).

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7 0
3 years ago
Ms. Wilson invested ​$34000 in two​ accounts, one yielding 8​% interest and the other yielding 9​%. If she received a total of​
Digiron [165]

Answer:

Amount invested at 8 % rate = x = $ 15000

Amount invested at 9 % rate = 34000 - x = 34000 - 15000 = $ 19000

Step-by-step explanation:

Total Amount = $ 34000

Let amount invested at 8 % rate = x

Amount invested at 9 % rate = $ 34000 - x

Total interest = $ 2910

2910 = \frac{x (8) (1)}{100} + \frac{(34000-x) (9) (1)}{100}

291000 = 8 x + 306000 - 9 x

x = 306000 - 291000

x = 15000

So  amount invested at 8 % rate = x = $ 15000

Amount invested at 9 % rate = 34000 - x = 34000 - 15000 = $ 19000

3 0
3 years ago
Pls help me.<br>proove it ^​
aivan3 [116]

Answer:

We verified that a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]

Hence proved

Step-by-step explanation:

Given equation is a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]

We have to prove that a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]

That is to prove that LHS=RHS

Now taking RHS

\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]

=\frac{a+b+c}{2}[a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2]  (using (a-b)^2=a^2-2ab+b^2)

=\frac{a+b+c}{2}[2a^2-2ab+2b^2-2bc+2c^2-2ac]  (adding the like terms)

=\frac{a+b+c}{2}[2a^2+2b^2+2c^2-2ab-2bc-2ac]

=\frac{a+b+c}{2}\times 2[a^2+b^2+c^2-ab-bc-ac]

=a+b+c[a^2+b^2+c^2-ab-bc-ac]

Now multiply the each term to another each term in the factor

=a^3+ab^2+ac^2-a^2b-abc-a^2c+ba62+b^3+bc^2-ab^2-b^2c-abc+ca^2+cb^2+c^3-abc-bc^2-ac^2]

=a^3+b^3+c^3-3abc (adding the like terms and other terms getting cancelled)

=a^3+b^3+c^3-3abc =LHS

Therefore LHS=RHS

Therefore a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]

Hence proved.

8 0
4 years ago
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