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Rudik [331]
3 years ago
8

How do you factor the quadratic equation 3x^2 + 8x + 3 to find the roots?

Mathematics
2 answers:
34kurt3 years ago
5 0

Answer:

Step-by-step explanation:

Hope this helps u!!

Fudgin [204]3 years ago
4 0
The quadratic is not factorable, to figure this out use this formula (b^2-4ac) to figure out the discriminant. If the discriminant comes out to be a perfect square or equal to zero then we can say it’s factorable if it does not meet those requirements it’s not. In this problem it’s 8^2-4*3*3. Simplify and you get 64-36 which is equal to 28. 28 is not a perfect square or equal to zero therefore this is not factorable. If you need the root the quadratic formula would be a good option also look on the bottom if that’s what u need.

Hope this helps!
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If each of the numbers in the following data set were multiplied by 31, what
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the answer to  ur question is c

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Which of the following describes the translation of y = |x| to = |x +7|
mote1985 [20]

Answer:

a

Step-by-step explanation:

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Answer:

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6 0
3 years ago
the vertex of this parabola is at (2,-1). when the y-value is 0, the x-value is 5. what is the coefficient of the squared term i
timama [110]
We know that

case a)
the equation of the vertical parabola write in vertex form is
y=a(x-h)²+k,
where (h, k) is the vertex.

Using our vertex, we have:
y=a(x-2)²-1
We know that the parabola goes through (5, 0),

so
we can use these coordinates to find the value of a:
0=a(5-2)²-1
0=a(3)²-1
0=9a-1

Add 1 to both sides:
0+1=9a-1+1
1=9a
Divide both sides by 9:
1/9 = 9a/9
1/9 = a
y=(1/9)(x-2)²-1

the answer is
a=1/9


case b)
the equation of the horizontal parabola write in vertex form is
x=a(y-k)²+h, 
where (h, k) is the vertex.

Using our vertex, we have:
x=a(y+1)²+2, 
We know that the parabola goes through (5, 0),

so
we can use these coordinates to find the value of a:
5=a(0+1)²+2
5=a+2
a=5-2
a=3
x=3(y+1)²+2

the answer is
a=3

see the attached figure 

3 0
3 years ago
Simplify (2z^5)(12z^3)/4z^4
Sphinxa [80]

Answer:

6z^{4}

Step-by-step explanation:

Given in the question an expression,

\frac{ (2z^5)(12z^3)}{4z^4}

Step 1

Apply exponential "product rule"

x^{m}x^{n}=x^{m+n}

\frac{ 12(2)z^5)(z^3)}{4z^4}

\frac{ (24)z^5)(z^3)}{4z^4}

\frac{ 24(z^{(5+3)})}{4z^4}

\frac{ 24(z^{8})}{4z^4}

Step 2

Apply exponential " divide rule"

\frac{x^{m}}{x^{n}}=x^{m-n}

\frac{24/4(z^{8})}{z^4}

\frac{6(z^{8})}{z^4}

\frac{6(z^{8-4})}{1}

6z^{4}

5 0
3 years ago
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