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3 years ago

Which equations have a greater unit rate than the rate represented in the table.

Mathematics

1 answer:

love history [14]3 years ago

5 0

Answer:

None. All of them equal to 1.5?

If its not none- then maybe its X:Y.. i.dk but when i divided all of them, they each gave me the answer of 1.5.

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Find the slope of the line containing the pair of points.<br> (6,5) and (0,0)

GarryVolchara [31]

Answer: 5/6

Step-by-step explanation:

Use slope formula of y2-y1/x2-x1. So here it doesn’t matter which is which so I did (6,5) x1 and y1 and (0,0) x2 and y2

So plugging it in it would be 0-5/0-6 = -5/-6 which simplifies to 5/6

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1 year ago

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pentagon [3]

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3 years ago

Point G(−7, 4) is translated using the rule (x+10, y−6). What is the x-coordinate of G′ ?

den301095 [7]

Answer:

x = 3

Step-by-step explanation:

The translation rule (x + 10, y - 6) means add 10 to the original x- coordinate and subtract 6 from the original y- coordinate, that is

G(- 7, 4) → G'(- 7 + 10, 4 - 6) → G'(3, - 2)

6 0

3 years ago

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Rzqust [24]

Factors and Multiples. But they both involve multiplication: Factors are what we can multiply to get the number. Multiples are what we get after multiplying the number by an integer (not a fraction).

3 0

4 years ago

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A wind turbine is located at the top of a hill where the wind blows steadily at 12 m/s, and stands 37 m tall. The air then exits

marta [7]

Answer:

Power generated = 4.315 kW

Step-by-step explanation:

Given,

speed of wind when enters into the turbine, V = 12 m/s

speed of wind when exits from the turbine, U = 9 m/s

mass flow rate of the wind, m = 137 kg/s

According to the law of conservation of energy

Energy generated = change in kinetic energy

Since, the air is exiting at same elevation. So, we will consider only kinetic energy.

Energy generated in one second will be given by,

$E\ =\ \dfrac{1}{2}.mV^2-\dfrac{1}{2}.m.U^2$

$=\ \dfrac{1}{2}\times 137\times (12)^2-\dfrac{1}{2}\times 137\times (9)^2$

=4315.5 J

= 4.315 kJ

So, energy is generated in one second = 4.315 kJ

Power generated can be given by,

$P\ =\ \dfrac{\textrm{energy generated}}{time}$

And the energy is generated is already in per second so power generated will be 4.315 kW.

8 0

3 years ago