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3 years ago

What are the solutions to the equation x2 +3x-4=0

2 answers:

4 0

$x^{2} +3x-4=0$

You wan't to get it into a format of (x+a)(x+b)=0
where a+b = 3 (the one from the 3x)
and where a*b= -4 (from the last -4)

(x+4)(x-1)=0
See how 4+(-1) = 3 and (4)(-1) = -4

now find what values of x would make the equation equal 0
x=-4
(-4+4)(-4-1) = (0)(-5) = 0

x=1
(1+4)(1-1) = (5)(0) = 0

frutty [35]3 years ago

4 0

X^2+3x-4=0
(x-1)(x+4)=0
x-1=0 x+4=0
x=1 x=-4
The solutions are 1 and -4.

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konstantin123 [22]

Answer:

a = 84/14

Step-by-step explanation:

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1 year ago

A 13 and one fourth inch board is cut from a 38 and three fourths inch board. The saw cut takes three eighths inch. How much of

gogolik [260]

Answer:

There is 25 1/8 inches left

Step-by-step explanation:

We have a board that is 38 3/4 inches

We cut away 13 1/4

38 3/4

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25 2/4

The saw takes 3/8

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Get a common denominator of 8 2/4 * 2/2 = 4/8

25 4/8

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25 1/8

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3 years ago

Read 2 more answers

Can someone pls help me out on the middle one pls ?:(

Sauron [17]

Answer:

77 C

Step-by-step explanation:

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5 0

2 years ago

Part 4: Use the information provided to write the vertex formula of each parabola.

sergey [27]

Answer: 1. x = (y - 2)² + 8

$\bold{2.\quad x=-\dfrac{1}{2}(y-10)^2}+1$

3. y = 2(x +9)² + 7

<u>Step-by-step explanation:</u>

Notes: Vertex form is: y =a(x - h)² + k or x =a(y - k)² + h

(h, k) is the vertex
point of vertex is midpoint of focus and directrix: $\dfrac{focus+directrix}{2}$

$\bullet\quad a=\dfrac{1}{4p}$

p is the distance from the vertex to the focus

$focus = \bigg(\dfrac{-31}{4},2\bigg)\qquad directrix: x=\dfrac{-33}{4}\\\\\text{Since directrix is x, then the x-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{-31}{4}+\frac{-33}{4}}{2}=\dfrac{\frac{-64}{4}}{2}=\dfrac{-16}{2}=-8\\\\\text{The y-value of the vertex is given by the focus as: 2}\\\\\text{vertex (h, k)}=(-8,2)$

Now let's find the a-value:

$p=focus-vertex\\\\p=\dfrac{-31}{4}-\dfrac{-32}{4}=\dfrac{1}{4}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{1}{4})}=\dfrac{1}{1}=1$

Now, plug in a = 1 and (h, k) = (-8, 2) into the equation x =a(y - k)² + h

x = (y - 2)² + 8

***************************************************************************************

$focus = \bigg(\dfrac{1}{2},10\bigg)\qquad directrix: x=\dfrac{3}{2}\\\\\text{Since directrix is x, then the x-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{1}{2}+\frac{3}{2}}{2}=\dfrac{\frac{4}{2}}{2}=\dfrac{2}{2}=1\\\\\text{The y-value of the vertex is given by the focus as: 10}\\\\\text{vertex (h, k)}=(1,10)$

Now let's find the a-value:

$p=focus-vertex\\\\p=\dfrac{1}{2}-\dfrac{2}{2}=\dfrac{-1}{2}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{-1}{2})}=\dfrac{1}{-2}=-\dfrac{1}{2}$

Now, plug in a = -1/2 and (h, k) = (1, 10) into the equation x =a(y - k)² + h

$\bold{x=-\dfrac{1}{2}(y-10)^2}+1$

***************************************************************************************

$focus = \bigg(-9,\dfrac{57}{8}\bigg)\qquad directrix: y=\dfrac{55}{8}\\\\\text{Since directrix is y, then the y-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{57}{8}+\frac{55}{8}}{2}=\dfrac{\frac{112}{8}}{2}=\dfrac{14}{2}=7\\\\\text{The x-value of the vertex is given by the focus as: -9}\\\\\text{vertex (h, k)}=(-9,7)$