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## 11th Grade Math

Ian Said:

How do you divide this rational expression (**11th**

**grade**

**math**)?

We Answered:

(4m + 8 / 3n - 3) / (2m + 6 / 7n - 7)<=> [ (4m+8) * (7n-7) ] / [ (3n-3) * (2m+6) ]

<=> [ 4(m+2)*7(n-1) ] / [ 3(n-1)*2(m+3) ]

<=> 28(m+2) / 6(m+3)

Are you sure it was 4m+8 and 2m+6 :) ?

Don Said:

How do you simplify this question (**11th**

**grade**

**math**)?

We Answered:

3x^2 + 9x - 30 (factor out the 3)3(x^2 + 3x - 10) (factor)

3(x - 2)(x + 5)

x^2 + 7x +10 (factor)

(x + 2)(x + 5)

3(x - 2)(x + 5)

(over)

(x + 2)(x+5) (the (x+5) is canceled out)

your answer is 3(x - 2) divided by (x + 2)

Ricky Said:

**11th**

**Grade**

**Math**Question (quadratic formula, solving) Help Please?

We Answered:

Let "x" be the width of the border. We know the outside length, L, and width, W, of the border is given by:L = 7 + 2x

W = 5 + 2x

The difference in area between the lawn-plus-border and lawn is:

A = LW - (7)(5) = 6.25

(7 + 2x)(5 + 2x) - 35 = 6.25

35 + 14x + 10x + 4x² - 35 = 6.25

4x² + 24x = 6.25

4x² + 24x - 6.25 = 0

The roots of the equation are found from the quadratic formula where, from the above equation, a = 4, b = 24, and c = -6.25:

x = {-b ± ?[b² - 4ac]}/2a = {-24 ± ?[(24)² - 4(4)(-6.25)]}/2(4) = -3 ± (1/8)?676

x = -6.25, 0.25 m

Since the border cannot have a negative length, the solution is x = 0.25 m

Check:

[7 + 2(0.25)][5 + 2(0.25)] - 35 = 6.25

Georgia Said:

How do I complete the square in this expression?**11th**

**Grade**

**Math**Help?

We Answered:

1. 4x^2+2x-12. 4x^2+2x=1 move the one over

3. x^2+1/2x=1/4 divide by four to remove coefficient.

4. x^2+1/2x+1/16x=1/4+1/16 divide your B term by 2 and square. Add to both sides

5. (x+1/4)^2 = 5/16 convert left side to squared form

6. x+1/4 = +or-?5/4 sq. root both sides.

7. x= -1/4+?5/4 and x= -1/4-?5/4 answer.

Juanita Said:

What are**11th**

**grade**studets learning in

**math**?

We Answered:

Trigonometry, Probability, Algebra, Arithmetic, Series and Sequences, Locus, Differentiation, GeometryAlbert Said:

How do u solve Equations in**11th**

**grade**

**math**?

We Answered:

1) Add like terms 8x+2x=10xYour equation should look like this

10x=15x-10

Subtract both sides by 15x

10x-15x=15x-15x-10

Substitute

-5x=-10

Divide both sides by -5

-5x/-5=-10/-5

Solve for x

x=2

2) Add like terms 4y+1y=5y

Your equation should look like this

5y+1=7(y-1)

Distribute

5y+1=7y-7

Add both sides by 7

5y+1+7=7y-7+7

Substitute

5y+8=7y

Subtract both sides by 5y

5y-5y+8=7y-5y

Substitute

8=2y

Divide both sides by 2

8/2=2y/2

Solve for y

y=4

3(2z-5)=2z+13

Distribute

6z-15=2z+13

Add both sides by 15

6z-15+15=2z+13+15

Substitute

6z=2z+28

Subtract both sides by 2z

6z-2z=2z+28

Substitute

4z=28

Divide both sides by 4

4z/4=28/4

Solve for z

z=7

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

Checking

1)8(2)+2(2)=15(2)-10

16+4=30-10

20=20

2) 4(4)+4+1=7(4-1)

16+4+1=7(3)

21=21

3(14-5)=14+13

3(9)=27

27=27

Hope that helps!

Linda Said:

What is the**11th**

**grade**

**math**course called MBF3C in Ontario?

We Answered:

the grade 11 math course is:Functions (in ontario.)

and there are three grade 12 ones :

Advanced Functions

Calculus

Data Management

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