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Lorico [155]
3 years ago
5

Ratio And Proportion

Mathematics
1 answer:
Allushta [10]3 years ago
4 0

Answer:

11200 for 7 engines for a total of 80 hours

Step-by-step explanation:

You have 8 engines burning 12000 .for 10 hours.

Divide 10 hours into 12000=1200 PER hour.

1200 ÷8 engines = 140 per hour per engine.

140 × # of hours (12)=1680 for 12 hours per engine.

# of engines × 1680 for 12 hour/per engine

5×1680 = 8400 of fuel for 12 hours/5 engine

Then you have 2 that runs 10 hours

2 × 140 per hour per engine =280 × 10 =2800 for 2 engines for 10 hours

8400. + 2800= 11200 of fuel for 6 engines

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The graph of $y = ax^2 + bx + c$ is shown below. Find $a \cdot b \cdot c$. (The distance between the grid lines is one unit.)
Savatey [412]

Answer:

a\cdot b\cdot c=\frac{15}{4}

Step-by-step explanation:

Vertex is the minimum or maximum point of parabola

Vertex of parabola is (h,k)

Therefore, from given graph (-3,-2) is the lowest point.

Vertex of parabola is at (-3,-2).

Standard equation of parabola

y-k=a(x-h)^2

Substitute the values

y-(-2)=a(x-(-3))^2=a(x+3)^2

y+2=a(x+3)^2

(-1,0) lies on the parabola.

Therefore, it satisfied the equation of parabola.

0+2=a(-1+3)^2=4a

a=2/4=1/2

Now, using the value of a

y+2=1/2(x+3)^2=1/2(x^2+6x+9)

y+2=\frac{1}{2}x^2+3x+\frac{9}{2}

y=\frac{1}{2}x^2+3x+\frac{9}{2}-2

y=\frac{1}{2}x^2+3x+\frac{9-4}{2}

y=\frac{1}{2}x^2+3x+\frac{5}{2}

By comparing with

y=ax^2+bx+c

We get

a=\frac{1}{2}, b=3, c=5/2

a\cdot b\cdot c=\frac{1}{2}\times 3\times \frac{5}{2}

a\cdot b\cdot c=\frac{15}{4}

7 0
2 years ago
What is the solution set?<br> O (0-2)<br> O (2.0)<br> O (7,0)<br> O (5,3)
muminat

Answer:

D. (5,3)

Step-by-step explanation:

The solution is (x,y). so in this case, you find the point where both arrows meet.

on the x-axis, which is positive right 5, and positive 3 up, on the y-axis

3 0
2 years ago
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cestrela7 [59]

Answer:

She only can play at most 4 chess games

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3 years ago
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Step-by-step explanation:

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The time shown on the clock is 11:05 starting at this time approximately what time will it be when the hands from an obtuse angl
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