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Wittaler [7]
3 years ago
12

What is the quotient and remainder of 32 divided by 6

Mathematics
1 answer:
bija089 [108]3 years ago
7 0

Answer:

5.33333333333333

Step-by-step explanation:

32 divided by 6

6 fits 5 times into 32

6*5 = 30

2 is left, 2 is 1/3 of 6.

1/3 = 0.33333333333

So, its 5.333

You might be interested in
Roger served 5/8 pound of cracker wich was 2/3 of the entire box what was the weight of the crackers originally in the box
horrorfan [7]
Answer:
original weight of the box = 15 / 16 pounds

Explanation:
Assume that the original amount in the box is x.
We are given that:
5/8 pounds represent 2/3 of the total amount (x).
This can be translated into the following equation:
(2/3) x = 5 / 8

Now, we will solve for x as follows:
(2/3) x = 5 / 8
Multiply both sides by 24 to get rid of the denominators as follows:
(2/3) x * 24 = (5 / 8) * 24
16 x = 15
Divide both sides by 16 to isolate the x as follows:
x = 15 / 16 pounds

Hope this helps :)

8 0
3 years ago
I need some one to do this fast!!!
Kruka [31]
It’s 8 u cross multiply 7 and 1 then 1 and 56 and then divide 56 by 7
7 0
3 years ago
8,264 is divisible by 3? True or false?
mars1129 [50]
Number\ is\ divisible\ by\ 3\ if\ the\ sum\ of\ the\ digits\ is\ divisible\ by\ 3.\\\\8,264\\\\8+2+6+4=20-
is\ not\ divisible\ by\ 3.\\\\8,264\ is\ divisible\ by\ 3-FALSE
7 0
3 years ago
Read 2 more answers
What is the answer to this equation <br> 40-32/8+5•-2
KiRa [710]

Answer:26

Step-by-step explanation

7 0
3 years ago
Tan(2 sin^-1 0.4)<br> Find the Exact value
stiv31 [10]

Answer:

tan(2u)=[4sqrt(21)]/[17]

Step-by-step explanation:

Let u=arcsin(0.4)

tan(2u)=sin(2u)/cos(2u)

tan(2u)=[2sin(u)cos(u)]/[cos^2(u)-sin^2(u)]

If u=arcsin(0.4), then sin(u)=0.4

By the Pythagorean Identity, cos^2(u)+sin^2(u)=1, we have cos^2(u)=1-sin^2(u)=1-(0.4)^2=1-0.16=0.84.

This also implies cos(u)=sqrt(0.84) since cosine is positive.

Plug in values:

tan(2u)=[2(0.4)(sqrt(0.84)]/[0.84-0.16]

tan(2u)=[2(0.4)(sqrt(0.84)]/[0.68]

tan(2u)=[(0.4)(sqrt(0.84)]/[0.34]

tan(2u)=[(40)(sqrt(0.84)]/[34]

tan(2u)=[(20)(sqrt(0.84)]/[17]

Note:

0.84=0.04(21)

So the principal square root of 0.04 is 0.2

Sqrt(0.84)=0.2sqrt(21).

tan(2u)=[(20)(0.2)(sqrt(21)]/[17]

tan(2u)=[(20)(2)sqrt(21)]/[170]

tan(2u)=[(2)(2)sqrt(21)]/[17]

tan(2u)=[4sqrt(21)]/[17]

7 0
3 years ago
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