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3 years ago

Solving QE's by factoring

Mathematics

1 answer:

gogolik [260]3 years ago

7 0

Answer:

The factored form of this is (x - 7)(x -4)

Step-by-step explanation:

In order to factor quadratic equaitons with a 1 as the first coefficient, we look for two numbers that multiply to the last number (28), but add up to the middle number (-11). To do this, we often list the factors of the last number.

Factors of 28

1, 28

-1, -28

2, 14

-2, -14

4, 7

-4, -7

The numbers -4 and -7 satisfy this requirement and are therefore the numbers we want. Now we stick them into parenthesis along with x terms in the front like this:

(x - 7)(x -4)

And that is the fully factored form.

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Solve<br> 3x^2= 147<br> x=<br> X=

marin [14]

Answer:

$3 {x}^{2} = 147 \\ or \: x^{2} = \frac{147}{3} \\ or \: x = \sqrt{49} \\ x = 7$

5 0

2 years ago

Read 2 more answers

how many 1/3 inch cubes does it take to fill a box with width 2 2/3 inches length 3 1/3 inches and height 2 1/3 inches

Alona [7]

Answer:

560/9 or around 63 cubes

Step-by-step explanation:

2 2/3 -> 8/3

3 1/3 -> 10/3

2 1/3-> 7/3

7/3 x 8/3= 56/9

56/9 x 10/3 = 560/27

560/27 divided by 1/3

= 560/9

8 0

2 years ago

Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -

GaryK [48]

Consider all parabolas:

$y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.$

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

$y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.$

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

$y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.$

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

$y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.$

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

$y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.$

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

$y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.$