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ikadub [295]
2 years ago
7

7. The length of a rectangle is 12 in. and the perimeter is 56 in. Find the width of the rectangle. A. 22 in. B. 16 in. C. 32 in

. D. 21 in
Mathematics
1 answer:
Eduardwww [97]2 years ago
5 0
I believe the answer would be b. I think this is because you know that the perimeter is the length times 2 and the width times 2 added together. If you do 12 times 2 you get 24. Then you subtract this from 56 because 56 is the perimeter of the whole thing. Then after this you get 32.Then divide 32 by 2 to find ONE of the widths.
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Please help me
galina1969 [7]

Answer:

Your answer is.....

In every big pieces there will be 3 * 1/3 pieces.

So total 12 of 1/3 pieces

Mark it as brainlist answer . follow me for more answer.

Step-by-step explanation:

7 0
2 years ago
Hello can someone help me simplify 4) and 5)? Thank you and please include steps :)
Leni [432]
Alrighty

remember some rules
(ab)^c=(a^c)(b^c)
and
x^{-m}=\frac{1}{x^m}
and
x^0=1 for all real values of x
and
(a^b)^c=a^{bc}
and
(\frac{a}{b})^c=\frac{a^c}{b^c}
and
(a/b)/(c/d)=(ad)/(bc)
and
(a^b)(a^c)=a^{b+c}
and
\frac{a^m}{a^n}=a^{m-n}
and don't forget pemdas
example: -x^m=-1(x^m) but (-x)^m=(-1)^m(x^m)

so

4.
(\frac{c^{-2}}{2})^{-2}=

\frac{(c^{-2})^{-2}}{2^{-2}}=

\frac{c^4}{\frac{1}{2^2}}=

\frac{c^4}{\frac{1}{4}}=

4c^4



5.

\frac{(-a)^4bc^5}{-a^2b^{-3}c^0}=

\frac{(-1)^4(a)^4bc^5}{-1(a^2)(\frac{1}{b^3}(1)}=

\frac{(1)a^4bc^5}{\frac{(-1)(a^2)}{b^3}}=

\frac{(a^4bc^5)b^3}{(-1)(a^2)}=

\frac{-a^4b^4c^5}{a^2}=

-a^2b^4c^5
3 0
3 years ago
-3(-4n + 30) = -30 need help​
DIA [1.3K]

First we will do -3 times everything in the parenthesis.

-3 times -4n = 12n

-3 times 30 = -90

SO,

12n - 90 = -30

       +90    +90

12n = 60

divide 12 on both sides;

n = 5

This is your answer,

IF THIS HELPED, PLS GIVE ME BRAINLIEST!

6 0
2 years ago
No. 4<br> please help, what’s a forty proportional?
jok3333 [9.3K]

Answer:

a/b = c/d --> d = b*c/a

d is the forth proportional

a) d = 2*3/1 = 6

b) d = 3*4/0.5 = 24

c) d = b*5/a = 5*b/a


6 0
3 years ago
If cos Θ = negative 4 over 7, what are the values of sin Θ and tan Θ?
FinnZ [79.3K]
Given cos theta is equal to - 4/ 7 then we can conclude that theta is in the second and third quadrants. In this case, the other leg is equal to square root of (7^2 - 4^2 ) equal to square root of 33. In this case, sin theta can be equal to +- square root of 33 / 7 and tan theta is equal to +-square root of 33 / 4.
7 0
3 years ago
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