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Genrish500 [490]

3 years ago

6

A motor is 85.4 % efficient: that is, the output is 85.4 % of the input. What is the input if the motor delivers 10.75 hp. ( Rou

nd your answer to 1 decimal place.)

1 answer:

Ludmilka [50]3 years ago

5 0

Answer:

12.6 hp

Step-by-step explanation:

Represent the input by n. 10.75 hp

Then 0.85n = 10.75 hp, and n (the input) is --------------- = 12.6 hp

0.85

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Kim and Melody are selling popcorn buckets for a school fundraiser. Customers can buy small buckets or large buckets. Kim sold 3

Papessa [141]

Answer:

The cost of

A small bucket = x = $8

A Larger bucket = y = $13

Step-by-step explanation:

Let the cost of

A small bucket = x

A Larger bucket = y

Kim sold 3 small buckets and 14 large buckets of popcorn for a total of $206.

3x + 14y = 206... Equation 1

Melody sold 11 small buckets and 11 large buckets of popcorn for a total of $231.

11x + 11y = 231.... Equation 2

What is the cost for a small bucket and a large bucket of popcorn?

3x + 14y = 206... Equation 1

11x + 11y = 231 ....... Equation 2

We solve using Elimination method

Multiply Equation 1 by 11 and Equation 2 by 3 to Eliminate x

33x + 154y = 2266.... Equation 3

33x + 33y = 693....... Equation 4

We subtract Equation 4 from Equation 3

121y = 1573

y = 1573/121

y = $13

Solving for x

3x + 14y = 206... Equation 1

3x + 14 × 13 = 206

3x + 182 = 206

3x = 206 - 182

3x = 24

x = 24/3

x = $8

Hence, the cost of

A small bucket = x = $8

A Larger bucket = y = $13

4 0

2 years ago

7x + 4x <br><br> I need help plz

Firlakuza [10]

Answer:

11x

Step-by-step explanation:

They have the same variable so just add the two numbers

6 0

3 years ago

Which is a perfect square?<br> 0 5<br> 0 8<br> 36<br> 44

Liono4ka [1.6K]

Answer:

C) 36

Step-by-step explanation:

Because 6^2=6*6=36.

3 0

3 years ago

Read 2 more answers

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 30 ft/s. Its height

Crank

Answer:

a) h = 0.1: $\bar v = -11\,\frac{ft}{s}$ , h = 0.01: $\bar v = -10.1\,\frac{ft}{s}$ , h = 0.001: $\bar v = -10\,\frac{ft}{s}$ , b) The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

Step-by-step explanation:

a) We know that $y = 30\cdot t -10\cdot t^{2}$ describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity ( $\bar v$ ), measured in feet per second, can be done by means of the following definition:

$\bar v = \frac{y(2+h)-y(2)}{h}$

Where:

$y(2)$ - Position of the ball evaluated at $t = 2\,s$ , measured in feet.

$y(2+h)$ - Position of the ball evaluated at $t =(2+h)\,s$ , measured in feet.

$h$ - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:

$h = 0.1\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}$

$y(2.1) = 18.9\,ft$

$\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}$

$\bar v = -11\,\frac{ft}{s}$

$h = 0.01\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}$

$y(2.01) = 19.899\,ft$

$\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}$

$\bar v = -10.1\,\frac{ft}{s}$

$h = 0.001\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}$

$y(2.001) = 19.99\,ft$

$\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}$

$\bar v = -10\,\frac{ft}{s}$

b) The instantaneous velocity when $t = 2\,s$ can be obtained by using the following limit:

$v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h-10\cdot t^{2}-20\cdot t \cdot h-10\cdot h^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot h-20\cdot t\cdot h-10\cdot h^{2}}{h}$

$v(t) = \lim_{h \to 0} 30-20\cdot t-10\cdot h$

$v(t) = 30\cdot \lim_{h \to 0} 1 - 20\cdot t \cdot \lim_{h \to 0} 1 - 10\cdot \lim_{h \to 0} h$

$v(t) = 30-20\cdot t$

And we finally evaluate the instantaneous velocity at $t = 2\,s$ :

$v(2) = 30-20\cdot (2)$

$v(2) = -10\,\frac{ft}{s}$

The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

8 0

3 years ago

1/4 + 1/2 + x = - 3/4 what is x equal to

ludmilkaskok [199]

Answer:

-6/4 or -1 1/2

8 0

2 years ago