All categories

3 years ago

Rewrite the following expression 15+21 using the GCF and the distributive property

Mathematics

2 answers:

saul85 [17]3 years ago

7 0

Answer:

$3(5+7)$

Step-by-step explanation:

We are asked to rewrite 15+21 using the GCF and the distributive property.

We know that greatest common factor of 15 and 21 is 3.

$15=3\times 5$ and $21=3\times 7$ .

We can rewrite our given expression as:

$3\times 5+3\times 7$

Now, we will factor out 3 from our expression as:

$3(5+7)$

Therefore, after rewriting our given expression using the GCF and the distributive property we will get $3(5+7)$ .

avanturin [10]3 years ago

3 0

First, we want to find the GCF of 15 and 21, which is 3. Now we do 15 divided by 3 and 21 divided by 3, which gives us 5 and 7. To rewrite the expression, we take the GCF and put it on the outside of the parenthesis. Inside the parenthesis, we put 5 + 7. So it'll look like this:

3(5 + 7)

You might be interested in

Julie and her brother go to Atlanta to ride the SkyView Ferris wheel. It measures 200 feet in diameter.

worty [1.4K]

Answer:

Part A) Circumference

Part B) $C=\pi D$

Part C) The distance traveled in one rotation is 628.32 feet

Step-by-step explanation:

Part A) we know that

The distance around the circle is equal to the circumference.

The Ferris Wheel have a circular shape

To find out the distance around the Ferris Wheel you should use the circumference

Part B) What is the formula needed to solve this problem?

we know that

The circumference is equal to multiply the number π by the diameter of the circle

$C=\pi D$

Part C) What is the distance traveled in one rotation?

we know that

One rotation subtends a central angle of 360 degrees

The distance traveled in one rotation is the same that the circumference of the Ferris wheel

we have

$D=200\ ft$ ----> diameter of the Ferris wheel

substitute in the formula of circumference

$C=\pi (200)\\C=200\pi\ ft$

assume

$\pi=3.1416$

$C=200(3.1416)=628.32\ ft$

therefore

The distance traveled in one rotation is 628.32 feet

3 0

3 years ago

.....................ok

Dafna11 [192]

Answer:

d. 2

Step-by-step explanation:

i hope this helps :)

5 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

Sboob wohs dna ten.esaelpemetaD ot og FG evah i naC

emmainna [20.7K]

Answer:

Ummmm no thanks. I am not going there.

Step-by-step explanation:

6 0

3 years ago

Solve x²-x-30=0 using the factorisation method.

Furkat [3]

Answer:

Step-by-step explanation:

x²-x-30=0

x² - 6x + 5x - 30 = 0

(x² - 6x) + (5x - 30) = 0

x(x - 6) + 5(x - 6) = 0

(x + 5)(x - 6) = 0

x + 5 = 0 or x - 6 = 0

x = -5 or x = 6

8 0

3 years ago