Factories create breakers for the following question -15x-6.

If 4x^2 - 100 = 0, the roots of the equation are

ololo11 [35]

$4x^2-100=0 \ \ \ \ \ \ |+100 \\
4x^2=100 \ \ \ \ \ \ \ \ \ \ \ |\div 4 \\
x^2=25 \\
x=\pm \sqrt{25} \\
x=\pm 5 \\
x=-5 \hbox{ or } x=5$

The answer is (3) -5 and 5.

5 0

3 years ago

I WILL MARK BRAINLIEST!!! !! (picture included with ONE question ^^^^^) only part a

Taya2010 [7]

Answer:

A. y = 1.5m + 7

B. at 7 months the baby will weigh 17.5 pounds

C. Since the baby is born weighing 7 pounds, at 0 months he will weigh 7 pounds. This tells us that the y-intercept will be seven. And since the baby is gaining 1.5 pounds every month, this means that the slope will be 1.5 and m will be the x of our y = mx + b equation. For part B, we simply plug in 7 to m and solve by multiplying 1.5 and 7 to get 10.5, then adding 7 to get 17.5.

6 0

3 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:

1.

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

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

2.

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

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

3.

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

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

4.

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

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

5.

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

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

6.

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

When x=2, y=13, then the point (2,13) is vertex of this sixth parabola.

3 0

3 years ago

The squares of the first five whole numbers

antiseptic1488 [7]

Answer:

Square each number: 1 , 2 , 3 , 4 , 5:

1² = 1 * 1 = 1

2² = 2 * 2 = 4

3² = 3 * 3 = 9

4² = 4 * 4 = 16

5² = 5 * 5 = 25

~

4 0

3 years ago

Solve for x. x+3.2=−3.5 (Any help at all is greatly appreciated!)

Maksim231197 [3]

You have to subtract 3.2 from 3.5 to get x. 0.3+ 3.2=2.5 so x=0.3

7 0

3 years ago

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