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3 years ago

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Justin has 20 pencils 25 erasers and 40 paperclip he organize them into groups with the same number of items in each group all t

he items in a group will be at same type how many items can you put in each group

1 answer:

e-lub [12.9K]3 years ago

8 0

Find the GCF (Greatest common factor) of all the numbers.
There will be 4 pencils. 5 erasers. 8 paper clips. Hope I could help!
Btw- the GCF is 5.

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Which answer orders the number from least to greatest

zmey [24]

Answer:

a

Step-by-step explanation:

7 0

3 years ago

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

3 years ago

A cyclist travels at an average speed of 13 mph for 3 hours.<br> How far did she travel in miles?

alexgriva [62]

Answer:

39 miles

Step-by-step explanation:

We know that a cyclist travelled at an average rate of 13 miles per hour for 3 hours and are asked to how far she travelled in miles.

To solve, we need to take into mind that for three hours, the cyclist drove 13 miles per hour.

Meaning that for every hour, 13 miles were driven.

Therefore we need to multiply 13 by 3 to get our answer :

13 miles * 3hrs

39 miles

5 0

3 years ago

Simplify these expressions

Xelga [282]

$\frac{10a}{2} \\ \\ 5a \\ \\$

The answer is: 5a

-----------

$\frac{36b}{6b} \\ \\ \frac{36}{6} \\ \\ 6 \\ \\$

The answer is: 6

------------

$\frac{16c^2}{8c} \\ \\ 2c^{2-1} \\ \\ 2c^1 \\ \\ 2c \\ \\$

The answer is: 2c

------------

$\frac{18ab^2}{2b} \\ \\ 9ab^{2-1} \\ \\ 9ab^1 \\ \\ 9ab \\ \\$

The answer is: 9ab

------------

$\frac{40b^2c^2}{5bc} \\ \\ 8b^{2-1} c^{2-1} \\ \\ 8b^1c^1 \\ \\ 8bc^1 \\ \\ 8bc \\ \\$

The answer is: 8bc

-----------

$\frac{63ab^2c^3}{7b^2c^2} \\ \\ \frac{63ac^3}{7c^2} \\ \\ 9ac^{3-2} \\ \\ 9ac^1 \\ \\ 9ac \\ \\$

The answer is: 9ac

6 0

4 years ago

Haos dog ate 40 cns of dog food in 31 day. how many cans should hao buy to feed his dog in 6 days

Crank

The answer is 8. 40 / 31 = 1.29032258. Multiply that by 6 and you have 7.74193548. Round up because you can't buy 7.7 cans of dog food.

7 0

3 years ago