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horsena [70]
2 years ago
5

Which function has a vertex at the origin?

Mathematics
1 answer:
dlinn [17]2 years ago
6 0

Answer:

  (d)  f(x) = -x²

Step-by-step explanation:

For the vertex of the quadratic function to be at the origin, both the x-term and the constant must be zero. That is, the function must be of the form ...

  f(x) = a(x -h)² +k . . . . . . . . . . vertex form; vertex at (h, k)

  f(x) = a(x -0)² +0 = ax² . . . . . vertex at the origin, (h, k) = (0, 0)

Of the offered answer choices, the only one with a vertex at the origin is ...

  f(x) = -x² . . . . . a=-1

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Step-by-step explanation:

24 + 24 + 120 + 160 + 200

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What is 250,000 divided by 12
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How do you simplify <img src="https://tex.z-dn.net/?f=7%5Csqrt%5B4%5D%7B80pq%5E%7B2%7Dr%5E%7B8%7D%20%20%7D" id="TexFormula1" tit
kondor19780726 [428]

Answer:

14 ( 5)^{\frac{1}{4}} p^{\frac{1}{4}} q^{\frac{1}{2}} r^{2}

Step-by-step explanation:

The expression to simplify in this problem is

7\sqrt[4]{80pq^2r^8}

Which can be rewritten as

7(80pq^2r^8)^{\frac{1}{4}}

Or also as

7\cdot 80^{\frac{1}{4}} \cdot p^{\frac{1}{4}} \cdot (q^2)^{\frac{1}{4}} \cdot (r^8)^{\frac{1}{4}}

Now we can apply the following rule for the calculation of the power of a power:

(a^m)^n = a^{m\cdot n}

So we get:

7\cdot 80^{\frac{1}{4}} \cdot p^{\frac{1}{4}} \cdot q^{\frac{1}{2}} \cdot r^{2}

Which can therefore be rewritten as

7\cdot (2^4\cdot 5)^{\frac{1}{4}} \cdot p^{\frac{1}{4}} \cdot q^{\frac{1}{2}} \cdot r^{2}

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which can be finally rewritten as

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Step-by-step explanation:

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To find (f/g)(5), find f(5) and (g5) then divide the values.

f(5) = 7 + 4(5) = 27

g(5) = 1/2 (5) = 2.5

27/2.5 = 10.8

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