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3 years ago

I need the help!!! I don't rember how to do this. Explain if u can.

Mathematics

1 answer:

-Dominant- [34]3 years ago

5 0

1. 3/2x+4=-9

3/2x=-13 (Combine like terms)

(divide both sides by 3/2)

x=-26/3 or -8 2/3

2. 2(3x-7)+4x=26

6x-14+4x=26 (distribute the 2)

10x-14=26 (combine like terms)

10x=40 (add 14 to both sides)

(Divide both sides by 10)

x=4

3. 4-3x=5-6x-7

4-3x=-2-6x (combine like terms)

4=-2-3x (add 3x to both sides)

6=-3x (add 2 to both sides)

(divide both sides by -3)

x=-2

4. 11y-3=5

11y=8 (combine like terms)

(divide both sides by 11)

y=8/11

5. 5-3x<14

-3x<9 (combine like terms)

(divide both sides by -3) (flip the < to > because youre dividing by a negative)

x >-3 The graph is a vertical (up and down) line at -3... it is a dashed line and shade towards the postitive numbers

6. x/x+2=3/7

7x=3(x+2) (cross multiply)

7x=3x+6 (distribute the 3)

4x=6 (combine like terms)

(divide both sides by 4)

x=6/4 or 1 2/4 or 1 1/2

7. 1.5/36=x/24 (instead of using 1 1/2, I used 1.5 since it is easier to work with)

36x=36 (cross multiply)

(divide both sides by 36)

X=1

8. y=2x+5 3x-2y=10 (this is a substitution problem)

3x-2(2x+5)=10 (plug in y=2x+5 in the second equation)

3x-4x-10=10 (distribute the -2)

-1x=20 (combine like terms and add 10 to both sides)

(divide each side by -1)

x=-20

(plug in -20 for either equation... I used y=2x+5)

y=2(-20)+5

y=-40+5 (distribute the 2)

y=-35

10. 6x-3y=11 6x+3y=17 (this is an elmination problem) (-3y & +3y cancle each other out so they will not be in the following steps until the end)

6x=11 +6x=17

12x=28 (combine the X's and 11+17)

(divide both sides by 12)

X= 7/3

(plug in 7/3 in either equation... i used 6x+3y=17)

6(7/3)+3y=17 (plug in 7/3 for x)

14+3y=17 (distribute)

3y=3 (combine like terms)

(divide both sides by 3)

y=1

10. 3x+5y=22 4x+3y=11 (since you can't eliminate here, you have to multiply both equations by a number to be able to... I multiplied the first equation by -3 and the second by 5. I'll be starting at the point after the multiplication)

-9x-15y=-66 20x+15y=55 (-15y & +15y cancle out)

-9x=-66 20x=55

11x=-11 (combine like terms)

(divide both sides by 11)

x=-1

(to find the value of y, plug in -1 into either equation and solve... i used 3x+5y=22)

3(-1)+5y=22 (plug in)

-3+5y=22 (distribute)

5y=25 (combine like terms)

(divide both sides)

y=5

11. n^2+3n-12=6

n^2+3n-18=0 (subtract 6 from both sides)

(n+3)(n-6) (to solve these, set them equal to 0 and get n alone)

n= -3, n=6

12. 2n^2+5n+2

(2n^2+n)+(4n+2)

n(2n+1)+2(2n+1)

(2n+1)(n+2)

MORE ANSWERS TO COME... STILL NEED TO SOLVE THEM

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In the circle below, suppose m VUX = 152° and mZUVW = 77º. Find the following.

bezimeni [28]

Given:

In the circle, $m(\overarc{VUX})=152^\circ$ and $m(\angle MUV)=77^\circ$ .

To find:

The following measures:

(a) $m\angle VUX$

(b) $m\angle UXW$

Solution:

According to the central angle theorem, the central angle is always twice of the subtended angle intercepted on the same same arc.

$m(VUX)=2\times m\angle VWX$

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$\dfrac{152^\circ}{2}=m\angle VWX$

$76^\circ=m\angle VWX$

In a cyclic quadrilateral, the opposite angles are supplementary angles.

UVWX is a cyclic quadrilateral. So,

$m\angle VUX+m\angle VWX=180^\circ$ [Opposite angles of a cyclic quadrilateral]

$m\angle VUX+76^\circ=180^\circ$

$m\angle VUX=180^\circ-76^\circ$

$m\angle VUX=104^\circ$

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$m\angle UXW+m\angle UVW=180^\circ$ [Opposite angles of a cyclic quadrilateral]

$m\angle UXW+77^\circ =180^\circ$

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$m\angle UXW=103^\circ$

Therefore, $m\angle VUX=104^\circ$ and $m\angle UXW=103^\circ$ .

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Umnica [9.8K]

2(8x - 4) = 32 distribute through parenthesis
16x - 8 =32 add 8 on both sides to isolate x
16x = 40 divide by 16 on both sides to isolate x
x= 40/16 simplify, Divide 40 by 16
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