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rewona [7]
3 years ago
7

Kaylee mixes 3/4 gal and 1/6 gal of paint. She uses 3/4 of the paint. How much paint does she use in all?

Mathematics
1 answer:
anyanavicka [17]3 years ago
3 0

Answer: 1/6

Step-by-step explanation:

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Arisa [49]

Step-by-step explanation:

26 * 14.6= 379.6 miles

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Solve the system by substitution y=10x y=-3x-104
NISA [10]

Answer:  (-8, -80)

<u>Step-by-step explanation:</u>

y = 10x

y = -3x - 104

Substitute "y" with "10x" into the second equation, then solve for x:

10x = -3x - 104

13x = -104         <em>added 3x to both sides</em>

  x = -8             <em>divided both sides by 13</em>

Next, substitute -8 for x into the first equation to solve for y:

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8 0
3 years ago
Please assist me with these problems​
Alex73 [517]

Answer:

1) $350

2) 55°

3) $300

Step-by-step explanation:

1) The total cost is the cost of the appointment plus the cost of the repair work.  The appointment costs $50.  The repair work costs $120 per 2 hours.  So the total cost is:

C = 50 + (120/2) t

C = 50 + 60t

For 5 hours of work, the cost is:

C = 50 + 60(5)

C = 350

It costs $350.

2) The temperature starts at 70°.  It drops 10° in 4 hours.  So the temperature after t hours is:

T = 70 + (-10/4) t

T = 70 − 2.5t

After 6 hours:

T = 70 − 2.5(6)

T = 55

The temperature is 55°.

3) Jennie's total pay is her weekly pay plus her commissions.  If her commission is x% of her sales, then her pay is:

P = 250 + (x/100) S

When her sales is $1000, her pay is $275.

275 = 250 + (x/100) 1000

275 = 250 + 10x

25 = 10x

x = 2.5

So her pay is:

P = 250 + (2.5/100) S

When S = $2000:

P = 250 + (2.5/100) 2000

P = 250 + 50

P = 300

Her total pay is $300.

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3 years ago
How can bar diagrams be used to model numerical expressions
vesna_86 [32]
A bar diagram can be used to represent algebraic expressions or to give examples.
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3 years ago
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Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necess
Zinaida [17]

since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill

\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2

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