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4 years ago

Joey buys 3 gallons of paint to paint his bedroom. He used 34 of the paint to paint the walls. How much

Mathematics

2 answers:

GrogVix [38]4 years ago

7 0

Answer:

11.3

Step-by-step explanation:

divide my dude

Anit [1.1K]4 years ago

6 0

Answer:

34 of the paint? is it 3/4 or just 34?

Step-by-step explanation:

You might be interested in

5 times the sum of 40 and 80. what is the numerical expression that would represent the phrase?

vlabodo [156]

<h2>Answer:</h2>

<u>The expression is</u><u> 5(40+80)</u>

<h2>Step-by-step explanation:</h2>

The sum of 40 and 80 can be written as

(40+80)

and when we take the 5 times of the sum

we can write as

5(40+80)

3 0

3 years ago

Prove algebraically that 0.5 recurring = 5/9

weeeeeb [17]

Answer:

see explanation

Step-by-step explanation:

We require 2 equations with the recurring part after the decimal point.

Let x = 0.555..... → (1)

Multiply both sides by 10, then

10x = 5.555..... → (2)

Subtract (1) from (2) thus eliminating the recurring decimal

(10x - x) = (5.5555 - 0.5555 ), that is

9x = 5 ( divide both sides by 9 )

x = $\frac{5}{9}$

5 0

3 years ago

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

A stadium has 45,000 seats. Seats sell for $28 in Section A, $24 in Section B, and $20 in Section C. If section C held 300 fewer

White raven [17]

Answer:

Let's define the variables:

A = number of seats in section A.

B = number of seats in section B.

C = number of seats in section C.

We have the equations:

A + B + C = 45,000.

C - 300 = B/2

A*$28 + B*$24 + C*$20 = $1,139,200

This is a system of equations, the first step to solve this is to isolate one variable in one of the equations, and then replace it in the others.

I will isolate C in the second equation:

C = B/2 + 300.

Now let's replace this in the other two equations:

A + B + B/2 + 300 = 45,000

A*$28 + B*$24 + (B/2 + 300)*$20 = $1,139,200

Let's simplify these equations:

A + B*(3/2) = 44,700

A*$28 + B*$34 + $6,000 = $1,139,200

Now let's isolate A in the first equation:

A = 44,700 - B*(3/2)

Let's replace this in the other equation:

(44,700 - B*(3/2))*$28 + B*$34 + $6,000 = $1,139,200

Now let's solve this for B.

-B*$8 + $1,252,200 = $1,139,200

-B*$8 = $1,139,200 - $1,252,200 = -$113,000

B = $113,000/8 = 14,125

Now we can replace that in the equations:

A = 44,700 - B*(3/2) = 44,700 - 14,125*(3/2) = 23,512.5, that we should round up to 23,513.

And C = B/2 + 300 = 7362.5

As we rounded the previous one up, we should round this one down to 7362.

3 0

3 years ago

In 7th grade there are 58 students with red-hair and 609 students with another color hair. The numbers of red-haired students in

mr_godi [17]

In 7th grade:

red-hair students =58

another color hair students =609

In 8th grade:

red-hair students =60

Let's assume

another color hair students =x

now, we are given

The numbers of red-haired students in 8th grade are proportional to the numbers in 7th

so,

$\frac{60}{58} =\frac{x}{609}$

now, we can solve for x

we get

$x=609*\frac{60}{58}$

we get

$x=630$

so, students have another color hair in 8th grade is 630.........Answer

5 0

3 years ago