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3 years ago

13

Okay, so lets say a teacher wants to divide her class of 30 students into 10 groups, but not necessarily of equal size. What are

some of her choices?

2 answers:

Phantasy [73]3 years ago

7 0

Well, she could do 30/10 which equals 3. So she could do 10 groups with 3 students in each group.

nata0808 [166]3 years ago

7 0

Simple, just pick a random number,like 2 or somthing.

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Jaylen bed is 6 feet long and his desk is 3 feet long. he wants to leave a two foot space between his bed and desk. if there is

Montano1993 [528]

No, he would need 132 inches

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2 years ago

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1. Use f(x)and g(x) to find f(x) * g(x).<br> f(x) = -x + 5 and g(x) = 3x + 2

iris [78.8K]

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2 years ago

The quotient of (x^3+3x^2-4x-12)/(x^2+5x+6)

Elis [28]

Answer: The quotient is (x-2).

Step-by-step explanation:

Since we have given that

$f(x)=(x^3+3x^2-4x-12)\\\\and\\\\g(x)=x^2+5x+6\\\\So,\ \frac{\left(x^3+3x^2-4x-12\right)}{\left(x^2+5x+6\right)}$

Now, we have to find the quotient of the above expression.

So, here we go:

$Factorise\ (x^3+3x^2-4x-12)\\\\=\left(x^3+3x^2\right)+\left(-4x-12\right)\\\\=-4\left(x+3\right)+x^2\left(x+3\right)\\\\=\left(x+3\right)\left(x^2-4\right)$

Now, we will divide the above simplest form with g(x):

$\frac{\left(x+3\right)\left(x^2-4\right)}{\left(x+2\right)\left(x+3\right)}\\\\=\frac{x^2-4}{x+2}\\\\=\frac{\left(x+2\right)\left(x-2\right)}{x+2}\ using\ (a^2-b^2)=(a+b)(a-b)\\\\=x-2$

Hence, the quotient is (x-2).

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3 years ago

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W = 3x + 7y solve for y

mario62 [17]

Answer:

The value of the equation $y=\frac{W-3x}{7}$ .

Step-by-step explanation:

Consider the provided equation.

$W = 3x + 7y$

We need to solve the provided equation for y.

Subtract 3x from both side.

$W-3x= 3x-3x+ 7y$

$W-3x=7y$

Divide both sides by 7.

$\frac{7y}{7}=\frac{W-3x}{7}$

$y=\frac{W-3x}{7}$

Hence, the value of the equation is $y=\frac{W-3x}{7}$ .

8 0

3 years ago

(3ab+2c)^3 what is the answer

svet-max [94.6K]

Answer:

$=27a^3b^3+54a^2b^2c+36abc^2+8c^3$

Step-by-step explanation:

$(3ab+2c)^3$

$=(3ab+2c)$ × $(3ab+2c)$ ×

$=(3ab+2c)(9a^2b^2+12abc+4c^2)$

$=(3ab)(9a^2b^2)+(3ab)(12abc)+(3ab)(4c^2)+(2c)(9a^2b^2)+(2c)(12abc)+(2c)(4c^2)$

$=27a^3b^3+36a^2b^2c+12abc^2+18a^2b^2c+24abc^2+8c^3$

$=27a^3b^3+54a^2b^2c+36abc^2+8c^3$

<em>hope this helps....</em>

5 0

2 years ago