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PilotLPTM [1.2K]
3 years ago
14

Jing bought 4 cans of paint and 1/2 of a pint of special paint additive formulated to reduce mildew. Before painting his house,

he divided the additive equally among the 4 cans of paint. How much additive did he put in each can?
Mathematics
2 answers:
Paul [167]3 years ago
6 0

Answer: There is 8 additive he can put in each can.

Step-by-step explanation:

Since we have given that

Number of cans = 4

Part of a pint of special paint add to reduce mildew = \dfrac{1}{2}

We need to find the additive that he put in each can.

So, the number of additive would be

4\div \dfrac{1}{2}\\\\=4\times 2\\\\=8

Hence, there is 8 additive he can put in each can.

xeze [42]3 years ago
5 0
Half of pint = 1 cup / 8oz
8 ÷ 2=2
2 oz per bucket
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Consider the following system of equations:
IrinaK [193]

Answer:

x = 6 ; y = -9

Step-by-step explanation:

8 x + 2 y = 30  ..........equ No 1

7 x + 2 y = 24......... equ No 2

8 x = 30 - 2y

∴ x =\frac{ 30 - 2 y}{8}

substituting the value of x in equ No 2

7 \frac{30 - 2 y}{8} + 2 y = 24

7 ( 30 -2 y) + 2 y × 8 = 24 × 8

7 × 30 - 14 y + 16 y = 192

210 + 2 y = 192

2 y = 192 - 210

2 y = - 18

∴ y = - 9

put y = -9 , 8 x = 30 - 2 y

        8 x = 30 - 2 ( -9)

    8 x = 30 - ( -18)

  8 x = 48

∴ x = 6

x = 6 ; y = -9

3 0
3 years ago
Carol used 114 cups of flour in 2 muffins. How many cups of flour are needed to make 12 similar muffins?
son4ous [18]

Answer:

7 1/2 cups

Step-by-step explanation:

If Carol uses 1 1/4 cups of flour for 2 muffins, then she will need to repeat this conversion 6 times since she wants to make identical 12 muffins in all.

So, 1 1/4 * 6 = 7 1/2

Therefore, she needs 7 1/2 cups of flour

Hope that helps! :)

7 0
3 years ago
A race car can go around the track 48 times in 8 minutes how many times can the race car go around per minute
BARSIC [14]

Answer:

Step-by-step explanation:

6

cause you take 48/8 equals 6

hope this heps:)

8 0
3 years ago
Please anyone answer me
ollegr [7]

Let's divide the shaded region into two areas:

area 1: x = 0 ---> x = 2

ares 2: x = 2 ---> x = 4

In area 1, we need to find the area under g(x) = x and in area 2, we need to find the area between g(x) = x and f(x) = (x - 2)^2. Now let's set up the integrals needed to find the areas.

Area 1:

A\frac{}{1}  = ∫g(x)dx =  ∫xdx =  \frac{1}{2}  {x}^{2}  | \frac{2}{0}  = 2

Area 2:

A\frac{}{2}  = ∫(g(x) - f(x))dx

= ∫(x -  {(x - 2)}^{2} )dx

=  ∫( - {x}^{2}  + 5x - 4)dx

= ( - \frac{1}{3}{x}^{3} +   \frac{5}{2} {x}^{2}  - 4x)   | \frac{4}{2}

= 2.67 - ( - 0.67) = 3.34

Therefore, the area of the shaded portion of the graph is

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3 0
3 years ago
I need help with number 19
inysia [295]
Let's start by understanding that the graph is a linear line. We have a common difference if 4.5 for every 10x we have more.

We can start by finding the gradient, using rise over run. Let's take (10, 4.5) and (20, 9)

Our rise is 9 - 4.5 = 4.5 and our run is 10. Then, our gradient becomes 4.5/10 = 0.45

Now, we can substitute points using the point-slope form.

y - 4.5 = 0.45(x - 10)
y - 4.5 = 0.45x - 4.5
Hence, our line becomes y = 0.45x and you can verify by substituting in points from the table.
4 0
3 years ago
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