Ask question

Ask question

All categories

3 years ago

6

In any difference 6th grade class the ratio of boys and two girls is 3:2 how many boys and how many girls could there be in one

class is there more than one possible answer

1 answer:

Artemon [7]3 years ago

4 0

Yes there can be more than one answer because it can go on forever
3 to 2
6 to 4
9 to 6
12 to 8
it can go on forever

You might be interested in

becky has the addresses of many people listed in her email book . she forwarded a copy of an article to all 5 of those people. i

Marianna [84]

P = 5

I am not sure if this is correct. However, P is used to represent the addresses and there are 5 addresses. Therefore p = 5.

8 0

2 years ago

NEED HELP ASAP (WILL GIVD BRAINELST)

Marizza181 [45]

Answer:

5

Step-by-step explanation:

The answer is 60/12 becuase we can simfly both of theese mixed numbers into an improper fraction, wich will be 4/3*15/4. We do the multiplication, and we get 60/12. 60/12 can be simplifed to 5 becuase 60 divided by 12 is 5

6 0

2 years ago

Newton has 28 cows Sally has 17 what is the sum?<br>

melamori03 [73]

Answer: 45

Step-by-step explanation:

Sum is addition so it would be

28+17=45

6 0

2 years ago

Read 2 more answers

Find the volume and area for the objects shown and answer Question

klio [65]

Step-by-step explanation:

You must write formulas regarding the volume and surface area of the given solids.

$\bold{\#1\ Rectangular\ prism:}\\\\V=lwh\\SA=2lw+2lh+2wh=2(lw+lh+wh)\\\\\bold{\#2\ Cylinder:}\\\\V=\pi r^2h\\SA=2\pi r^2+2\pi rh=2\pir(r+h)\\\\\bold{\#3\ Sphere:}\\\\V=\dfrac{4}{3}\pi r^3\\SA=4\pi r^2$

$\bold{\#4\ Cone:}\\\\V=\dfrac{1}{3}\pi r^2h\\\\\text{we need calculate the length of a slant length}\ l\\\text{use the Pythagorean theorem:}\\\\l^2=r^2+h^2\to l=\sqrt{r^2+h^2}\\\\SA=\pi r^2+\pi rl=\pi r^2+\pi r\sqrt{r^2+h^2}=\pi r(r+\sqrt{r^2+h^2})\\\\\bold{\#5\ Rectangular\ Pyramid:}\\\\V=\dfrac{1}{3}lwh\\\\$

$\\\text{we need to calculate the height of two different side walls}\ h_1\ \text{and}\ h_2\\\text{use the Pythagorean theorem:}\\\\h_1^2=\left(\dfrac{l}{2}\right)^2+h^2\to h_1=\sqrt{\left(\dfrac{l}{2}\right)^2+h^2}=\sqrt{\dfrac{l^2}{4}+h^2}=\sqrt{\dfrac{l^2}{4}+\dfrac{4h^2}{4}}\\\\h_1=\sqrt{\dfrac{l^2+4h^2}{4}}=\dfrac{\sqrt{l^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{l^2+4h^2}}{2}$

$\\\\h_2^2=\left(\dfrac{w}{2}\right)^2+h^2\to h_2=\sqrt{\left(\dfrac{w}{2}\right)^2+h^2}=\sqrt{\dfrac{w^2}{4}+h^2}=\sqrt{\dfrac{w^2}{4}+\dfrac{4h^2}{4}}\\\\h_2=\sqrt{\dfrac{w^2+4h^2}{4}}=\dfrac{\sqrt{w^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{w^2+4h^2}}{2}$

$SA=lw+2\cdot\dfrac{lh_1}{2}+2\cdot\dfrac{wh_2}{2}\\\\SA=lw+2\!\!\!\!\diagup\cdot\dfrac{l\cdot\frac{\sqrt{l^2+4h^2}}{2}}{2\!\!\!\!\diagup}+2\!\!\!\!\diagup\cdot\dfrac{w\cdot\frac{\sqrt{w^2+4h^2}}{2}}{2\!\!\!\!\diagup}\\\\SA=lw+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw}{2}+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw+l\sqrt{l^2+4h^2}+w\sqrt{w^2+4h^2}}{2}$

6 0

2 years ago

9. (3x + 4y)² – (3z) 2square<br>

rodikova [14]

Solution: (3x + 4y - 3z) x (3x + 4y + 3z)

5 0

3 years ago

Read 2 more answers