What is the LEAST common denominator for 6 and 8

Answer:

You do this by finding the area by each rectangle then the area of the quarter of the circle? (dont know what to call it) Then add it and round to the nearest tenth

Step-by-step explanation:

Answer:

d. 6.4

Step-by-step explanation:

d=√(1−5)2+(−3−2)2

d=√(−4)2+(−5)2

d=√16+25

d=√4–√1

d=6.403124

Answer:

[tex](y - 1)(x + y + 1)[/tex]

Step-by-step explanation:

[tex]y(x + y) - 1(1 + x)[/tex]

Lets expand it.

[tex] = > xy + {y}^{2} - x - 1[/tex]

Lets re-arrange it.

[tex] = > xy - x + {y}^{2} - 1[/tex]

Lets split [tex] {y}^{2} - 1[/tex]

So,

[tex] = > x(y - 1) + (y + 1)(y - 1)[/tex]

[tex] = > (y - 1)(x + y + 1)[/tex]

Answer:

hope it helps you have a great day keep smiling be happy stay safe

Answer:

-19f + 7

Step-by-step explanation:

group the fs

-19f + 7

is the simple expression

Answer:

it's A,B and D , Edge 2021

Step-by-step explanation:

Answer:

The person over me is right, the correct answers are A,B and D.

Step-by-step explanation: I got them right. Edge 2021.

Answer:

5

Step-by-step explanation:

3 1/8 : 8 x 3=25 new fraction

25/8 x 8/5 you have to do kcf Keep Change Flip

and then multiply

8x5=40 25x8=200 simply by 40

5/1= 5 hope it helped im not the best :)  

1) First find the number of buses

Equation:

total students ÷ 60 students per bus = number of buses taken

       340         ÷               60                 =     5 buses taken with 40 students left

2) Now find the number of cars taken

Equation:

students left ÷ 5 students per car = number of cars taken

        40        ÷                5               =         8 cars taken with 0 students left

So your answer is 8 cars and 5 buses.

I hope this helps!

-sincerelynini

Answer:

I think #2

Step-by-step explanation:

Given:

A caterpillar moves across the grass at a rate of [tex]4\dfrac{2}{3}[/tex] feet every 7 minutes.

To find:

The rate of moving caterpillar in feet per minute.

Solution:

We have,

Distance covered by caterpillar is 7 minutes = [tex]4\dfrac{2}{3}[/tex] feet

Distance covered by caterpillar is 1 minutes = [tex]\dfrac{4\dfrac{2}{3}}{7}[/tex] feet

                                                                        = [tex]\dfrac{\dfrac{4(3)+2}{3}}{7}[/tex] feet

                                                                        = [tex]\dfrac{14}{3\times 7}[/tex] feet

                                                                        = [tex]\dfrac{2}{3}[/tex] feet

Therefore, caterpillar move [tex]\dfrac{2}{3}[/tex] feet per minute.

Answer: 2.45

Step-by-step explanation:

I use pythagorean theorem after i found the total of leg A and leg B

Answer:

Here you go. It is really easy. All you have to do is translate the points. If it says translate 3 units up, you'd pick a point, and move it up 3 units. Then do it with all the others and BOOM! You did it! Try one of the problems yourself to make sure you understand what to do next time.

Have a nice day!

Answer:

6 i think

Step-by-step explanation:

Answer:

24/4 or 6

Step-by-step explanation:

3/4 / 1/8

flip the 1/8 and the sign

3/4x8/1

3x8=24

4x1=4

Step-by-step explanation:

[tex] - 4( - 6 + 2) [/tex]

[tex] - 4( - 4) = + 16[/tex]

Answer: A)-16

Step-by-step explanation:

Answer:

10/27 cups

Step-by-step explanation:

y = how much flour is needed

set up proportion of water to flour equals water to flour

[tex]\frac{3/5}{2/3}[/tex] = [tex]\frac{1/3}{y}[/tex]

cross-multiply:  3/5 y = 2/9

multiply each side by the reciprocal of 3/5

5/3 · (3/5)y = 2/9 · (5/3)

y = 10/27

Answer: 10/27 cups flour.

Step-by-step explanation:

Compare fractions 3/5 and 1/3:

[tex]\dfrac{3}{5}=\dfrac{3 \times 3}{5 \times 3} =\dfrac{9}{15} \\\\ \dfrac{1}{3}=\dfrac{1 \times 5}{3 \times 5} =\dfrac{3}{15}\\\\\dfrac{3}{15} < \dfrac{9}{15}[/tex]

1/3 is less than 3/5 because 3/15 is less than 9/15.

Find how many times 1/3 is less than 3/5:

[tex]\dfrac{3}{5} \div \dfrac{1}{3} =\dfrac{3}{5} \times \dfrac{3}{1}=\dfrac{9}{5}[/tex]

This means that flour should also be taken 9/5 times less:

[tex]\dfrac{2}{3} \div \dfrac{9}{5} =\dfrac{2}{3} \times \dfrac{5}{9} =\dfrac{10}{27}[/tex]

Answer:

Maybe M' (3, -2)

Step-by-step explanation:

Answer:

[tex] \huge \boxed{3 {x}^{2} + 5x - 4}[/tex]

In this equation the constant is -4.

2πr

2x22/7x2

4 x 22/7

88/7

=12.57142

Answer: 8 feet

Step-by-step explanation:

Answer:

1,200 cubic cm

Step-by-step explanation:

First, you separate it into two prisms, which gives you two different volume formulas.

14*20*2 = 560

20*4*8 = 640

You then add the two volumes together, which equals 1,200

Answer:

sorry wish i could help....

Step-by-step explanation:

Answer:

6(9-4x)

Step-by-step explanation:

Factor 6 out of 54-24x, or group and factor out the GCF. Hope this helps!

Answer:

0.5 X 2 X 3 = 3

0.5 X 3 X 3 = 4.5

8 X 4 = 32

32 + 4.5 + 3 = 39.5 square ft

Answer:

its b

Step-by-step explanation:

Just plain math. I was on this question

Answer:

d

Step-by-step explanation:

add them all up then divide by 5

Answer:

Alicia is 18years

Quentin is 14years

Rosalyn is 16years

Step-by-step explanation:

Let the age of Alicia be x

Let the age of Quentin be y

Let the age of Rosalyn be z

If the sum of the siblings’ ages, in years, is 48 then;

x+y+z = 48.... 1

If Rosalyn is two years older than Quentin then;

z =y+2 ..... 2

Also if Quentin is four years younger than Alicia then;

y = x-4 ... 3

From 3:

x = y+4 ... 4

Substitute 2 and 4 into 1 to have:

y+4+y+y+2 = 48

3y + 6= 48

3y = 48-6

3y= 42

y = 14

Since x= y+4

x = 14+4

x = 18

Since z = y+2

z = 14+2

z = 16year

This means that;

Alicia is 18years

Quentin is 14years

Rosalyn is 16years

Answer:

how many hours will she need to work in order to have enough money to go to the fair with her friends? $60/$7.5/h = 8h

Step-by-step explanation:

Answer:

8 hours

Step-by-step explanation:

60=7.50x

60 will represent how much money she needs to earn.

x= number of hours Paige needs to work

[tex]\frac{60}{7.50} =\frac{7.50x}{7.50}[/tex]

[tex]x=8[/tex]

Paige needs to work 8 hours in order to have enough money to go to the fair with her friends.