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MariettaO [177]
3 years ago
5

Simplify each expression

Mathematics
1 answer:
avanturin [10]3 years ago
4 0

-6x-4y+2+3x+12y-6=(-6x+3x)+(-4y+12y)+(2-6)\\\\=-3x+8y-4\\\\-15x-3y+16x+4y=(-15x+16x)+(-3y+4y)=x+y\\\\-6x-4y+2+3x-12y+6=(-6x+3x)+(-4y-12y)+(2+6)\\\\=-3x-16y+8\\\\2x+2y-6x-3+2y+6=(2x-6x)+(2y+2y)+(-3+6)\\\\=-4x+4y+3

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The graph below represents the solution set of which inequality?
natulia [17]

Answer:

option: B (x^2+2x-8) is correct.

Step-by-step explanation:

We are given the solution set as seen from the graph as:

(-4,2)

1)

On solving the first inequality we have:

x^2-2x-8

On using the method of splitting the middle term we have:

x^2-4x+2x-8

⇒  x(x-4)+2(x-4)=0

⇒ (x+2)(x-4)

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

x+2>0 and x-4

i.e. x>-2 and x<4

so we have the region as:

(-2,4)

Case 2:

x+2 and x-4>0

i.e. x<-2 and x>4

Hence, we did not get a common region.

Hence from both the cases we did not get the required region.

Hence, option 1 is incorrect.

2)

We are given the second inequality as:

x^2+2x-8

On using the method of splitting the middle term we have:

x^2+4x-2x-8

⇒ x(x+4)-2(x+4)

⇒ (x-2)(x+4)

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

x-2>0 and x+4

i.e. x>2 and x<-4

Hence, we do not get a common region.

Case 2:

x-2 and x+4>0

i.e. x<2 and x>-4

Hence the common region is (-4,2) which is same as the given option.

Hence, option B is correct.

3)

x^2-2x-8>0

On using the method of splitting the middle term we have:

x^2-4x+2x-8>0

⇒ x(x-4)+2(x-4)>0

⇒ (x-4)(x+2)>0

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

x+2>0 and x-4>0

i.e. x>-2 and x>4

Hence, the common region is (4,∞)

Case 2:

x+2 and x-4

i.e. x<-2 and x<4

Hence, the common region is: (-∞,-2)

Hence, from both the cases we did not get the desired answer.

Hence, option C is incorrect.

4)

x^2+2x-8>0

On using the method of splitting the middle term we have:

x^2+4x-2x-8>0

⇒ x(x+4)-2(x+4)>0

⇒ (x-2)(X+4)>0

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

x-2 and x+4

i.e. x<2 and x<-4

Hence, the common region is: (-∞,-4)

Case 2:

x-2>0 and x+4>0

i.e. x>2 and x>-4.

Hence, the common region is: (2,∞)

Hence from both the case we do not have the desired region.

Hence, option D is incorrect.




5 0
3 years ago
If the length of the side of a square is 3×-y what is the area of the square in terms of x and y
soldier1979 [14.2K]
(3x -y)(3x-y )= 9x^2 - 6xy + y^2
5 0
3 years ago
Find the volume of this figure and show your work
mario62 [17]

Answer:

10

Step-by-step explanation:

The answer is 10 because there are five stacks of 2 cubes. 5 x 2 = 10.

4 0
3 years ago
Find the measure of ∠C if
V125BC [204]
(32x+20) + (72x+20) + (52x+40) + (32x+10)=180 \\ 32x+72x+52x+32x +20+20+40+10 =180 \\ 188x+90=180 \\ -90  \\ 188x=90 \\ /188 \\ x=0.478&#10;
then 52(0.478)+40 \\ 24.856+40=64.856&#10;
∠C = 64.856
7 0
3 years ago
Read 2 more answers
360= 90 + (3x +12) + (4y+20) + y
ella [17]

answer:

x=238/3-5/3y,y

6 0
3 years ago
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