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3 years ago

Helppp me please IREADY

Mathematics

1 answer:

Morgarella [4.7K]3 years ago

7 0

Hello there! The answer is the first option, 31/55.

To solve this, we don't even need to do any math. Note that fractions with the same numerator and denominator will be equal to 1.

Knowing this and looking at our second and fourth options, 55/55 x 111 and 31/31 x 111, these problems are the same as 1 x 111, which results in 111, which is not less than, leaving us with the third and first options.

The third option is 55/31, meaning we have more than a whole, so we are multiplying by a number greater than 1, making our answer over 111 and this option not correct.

This leaves the first option as your answer!

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A rectangle has dimensions of 15 centimeters long and 16 meters wide what is the area of the rectangle

Brrunno [24]

240 is the answer to your question just multiply them

6 0

3 years ago

Gerry thinks that the points (4,2) and (−1,4) form a line perpendicular to a line with slope 4. Do you agree? Why

faust18 [17]

Answer:

The answer to your question is these lines are not perpendicular.

Step-by-step explanation:

Data

A (4, 2)

B (-1, 4)

slope = m = 4

Perpendicular lines mean that these lines cross and form an angle of 90°. Also, the slope of perpendicular lines is negative reciprocals.

Process

1.- Find the slope of the second line and compare it to the slope given.

slope = $\frac{y2 - y1}{x2 - x1}$

Substitution

slope = $\frac{4 - 2}{-1 - 4}$

Simplification and result

slope = $\frac{-2}{5}$

-2/5 is not a negative reciprocal of 4, so these lines are not perpendicular.

4 0

3 years ago

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

Please help with the following question

Mrrafil [7]

The generic equation of the line is:
y-yo = m (x-xo)
The slope is:
m = (y2-y1) / (x2-x1)
m = (- 2-0) / (3-0)
m = -2 / 3
We choose an ordered pair
(xo, yo) = (0, 0)
Substituting values:
y-0 = (- 2/3) (x-0)
Rewriting:
y = (- 2/3) x
Answer:
The equation of the line is:
y = (- 2/3) x

4 0

3 years ago

A rectangle has a width of 7. 9 ft and a perimeter of 27. 2 ft.

antiseptic1488 [7]

Answer:

The height would be 5.7 ft and the area would be 45.03

THIS IS A 2D SHAPE

7 0

2 years ago

Helppp me please IREADY​

Helppp me please IREADY