All categories

2 years ago

What is 4(m + 6 please answer fast

Mathematics

2 answers:

Hitman42 [59]2 years ago

5 0

Answer:

4m+24

Step-by-step explanation:

Thank you:)

olga55 [171]2 years ago

3 0

Hello there!

4(m+6)

Use the Distributive Property which states:

a(b+c) =ab+ac

4(m+6)

4m+24

That's the answer. Hope it helps! :)

Good luck

You might be interested in

You decided to go to the mall and find a sweatshirt that cost $25.80 on sale for 35% off. How much will you SAVE by buying the s

dusya [7]

Answer and Step-by-step explanation:

25.80*35/100=9.03

you will save 9.03

Hope I made you happy :)

6 0

3 years ago

I need help no ilnks please and thank you

Elza [17]

Answer:

yep

Step-by-step explanation:

nice

8 0

3 years ago

Please answer need for geometry

IrinaK [193]

Answer:

d. 8.38 ft

Step-by-step explanation:

arc(NM) = 180° - 100° = 80° (Angles on a straight line)

Length of arc = θ/360°*2πr

θ = 80°

r = 6 ft

Plug in the values

Length of arc(NM) = 80/360*2*π*6

= 8.3775804

= 8.38 ft

7 0

3 years ago

What is an equation of the line that passes through the points (4,-2) and (-1,3)

Paraphin [41]

Answer:

<h3>The slope is -1.</h3>

Step-by-step explanation:

To find an equation of the line that passes through the points, you have to use the slope formula.

Slope:

$\Rightarrow \sf{\dfrac{y_2-y_1}{x_2-x_1} }$

y2=3

y1=(-2)

x2=(-1)

x1=4

Solve.

3--2/-1-4=5/-5=-1

The slope is -1.

4 0

1 year ago

Read 2 more answers

Suppose the half-life of a material is 10 days. You have one 1 kg of the material today. How much of the material would you have

boyakko [2]

Answer:

0.50 kg of the material would be left after 10 days.

0.25 kg of the material would be left after 20 days.

Step-by-step explanation:

We have been given that the half-life of a material is 10 days. You have one 1 kg of the material today. We are asked to find the amount of material left after 10 days and 20 days, respectively.

We will use half life formula.

$A=a\cdot(\frac{1}{2})^{\frac{t}{h}}$ , where,

A = Amount left after t units of time,

a = Initial amount,

t = Time,

h = Half-life.

$A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{10}{10}}$

$A=1\text{ kg}\cdot(\frac{1}{2})^{1}$

$A=1\text{ kg}\cdot\frac{1}{2}$

$A=0.5\text{ kg}$

Therefore, amount of the material left after 10 days would be 0.5 kg.

$A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{20}{10}}$

$A=1\text{ kg}\cdot(\frac{1}{2})^{2}$

$A=1\text{ kg}\cdot\frac{1^2}{2^2}$

$A=1\text{ kg}\cdot\frac{1}{4}$

$A=0.25\text{ kg}$

Therefore, amount of the material left after 20 days would be 0.25 kg.

6 0

3 years ago