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Maslowich
3 years ago
15

A 12-pint bucket of paint costs $51.60. What is the price per quart?

Mathematics
1 answer:
My name is Ann [436]3 years ago
6 0

Answer: $8.60

Step-by-step explanation:

1 pint = 0.5 quart

12 pint = 6

51.6/6 = 8.6

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Solve the equation x1/3=2
NemiM [27]

Step-by-step explanation:

X1/3=2

X1=2×3

X1=5

x=5/1

4 0
3 years ago
-9 + y = 12 <br> How do you solve this?
Goshia [24]

Answer:

-9+y=12

y=12+9

=21

Ans: y=21

6 0
3 years ago
What is the volume of this rectangular prism? The length is 5 1/2. The width is 1 3/8. The height is 1 2/3.
natka813 [3]

Answer:

12  29 /48  in mixed number form,

≈ 12.6 as a rounded decimal

Step-by-step explanation:

Multiply all three numbers together for the volume.

3 0
3 years ago
A certain office supply store stocks 2 sizes of self-stick notepads, each in 4 colors: blue, green, yellow, or pink. The store p
liq [111]

Answer:  (C) 16

Step-by-step explanation:

Given : Number of sizes of self-stick notepads= 2

Number of colors = 4

Type 1 : The store packs the notepads in packages that contain 3 notepads of the same size.

i.e. By Fundamental principle of counting , we have

Number of different packages possible = Number of colors x No. of sizes

= 4\times2=8

Type 2: Notepads of the same size and of 3 different colors.

Choices for 3 different colors out of 4 : ^4C_3=\dfrac{4!}{3!(4-3)!}=4

Again by fundamental principle :

Number of different packages possible = No. of choices for choosing 3 different colors x No. of sizes

= 4\times2=8

From Type 1 and Type 2 , the total number of different packages of the types described above are possible =8+8= 16

Hence, the correct answer is option (c) 16.

4 0
3 years ago
Verify sin^4x − sin^2x = cos^4x − cos^2x is an identity
Inga [223]

Answer:

<h2>It's an identity</h2>

Step-by-step explanation:

\sin^4x-\sin^2x=\cos^4x-\cos^2x\\\\L_s=\sin^4x-\sin^2x=\sin^2x(\sin^2x-1)=\sin^2x(-\cos^2x)=-\sin^2x\cos^2x\\\\R_s=\cos^4x-\cos^2x=\cos^2x(\cos^2x-1)=\cos^2x(-\sin^2x)=-\sin^2x\cos^2x\\\\L_s=R_s\qquad\bold{CORRECT}

\text{Used:}\\\\\sin^2x+\cos^2x=1\Rightarrow\left\{\begin{array}{ccc}\cos^2x=1-\sin^2x\\\sin^2x=1-\cos^2x\end{array}\right\Rightarrow\left\{\begin{array}{ccc}-\cos^2x=\sin^2x-1\\-\sin^2x=\cos^2x-1\end{array}\right

4 0
3 years ago
Read 2 more answers
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