3 years ago

Find the quotient and remainder for52/8

Mathematics

1 answer:

Burka [1]3 years ago

8 0

The answer 48 with a remainder of 4. Hope this works.

You might be interested in

Simplify hariy fractions

borishaifa [10]

Um sounds nice i guess its kind of funny simplify hairy fractions

3 0

2 years ago

PLZZZZZ HELP ME Find the sum

mario62 [17]

Answer:

$27 \frac{2}{3}$

Step-by-step explanation:

Ok, so first substitute x for 4 in

"g(x) = 5x + 1", and x for 3 in

"k(x) = 2/x + 2x". Now you got:

g(4) = 5(4) + 1

k(3) = 2/(3) + 2(3)

Now you can solve each individually.

g(4) = 5 × 4 = 20

20 + 1 = 21

g(4) = 21

k(3) = 2 × 3 = 6

6 + 2/3 = 6 2/3

k(3) = 6 2/3

g(4) + k(3) = 21 + 6 2/3 = <u>27 2/3</u>

Hope this helps :)

7 0

2 years ago

Which expression is equivalent to [(3xy^-5)^3/(x^-2y^2)^-4]^-2?

Mademuasel [1]

Answer:- a.The given expression is equivalent to $\frac{x^{10}y^{14}}{729}$

Given expression:- $[\frac{(3xy^{-5})^3}{(x^{-2}y^2)^{-4}}]^{-2}$

$=[\frac{(3)^3x^3y^{-5\times3}}{x^{-2\times-4}y^{2\times-4}}]^{-2}.........(a^m)^n=a^{mn}$

$=[\frac{27x^3y^{-15}}{x^8y^{-8}}]^{-2}$

$=[27x^{3-8}y^{-15-(-8)}]^{-2}............\frac{a^m}{a^n}=a^{m-n}$

$=[27x^{-5}y^{-7}]^{-2}=(27)^{-2}(x^{-5})^{-2}(y^{-7})^{-2}.........(a^m)^n=a^{mn}$

$=\frac{1}{(27)^2}(x^{10}y^{14})=\frac{x^{10}y^{14}}{729}$

Thus a. is the right answer.

6 0

3 years ago

Read 2 more answers

Determine whether the given ordered pair is a solution to the given equation. <br> (0,-5) x+4y=-20

svetoff [14.1K]

yes

Substitute x and y into the

equation we have

0+4 .(-5)=-20

<=>-20=-20

satisfy the condition

6 0

3 years ago

P varies jointly with t and inversely with s. If p = 4 when t = 3 and s = 6, find p when t = -6 and s = 2.

WINSTONCH [101]

Answer:

Step-by-step explanation:

4 0

2 years ago