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Kruka [31]
2 years ago
14

Complete the Square Rewrite into Vertex Form by completing the square. 2x^2+ 20x – 1

Mathematics
1 answer:
Stolb23 [73]2 years ago
5 0
I believe the answer is : Y=2(x+5)^2-51
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Answer:

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Step-by-step explanation:

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Given X(0, 2), Y(-2, 2), and Z(-2, 0), find the coordinates of X', Y', and Z' after a dilation with scale factor -4. ​
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Answer:

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3 0
3 years ago
Use this information to answer parts A, B, C, and D.
ikadub [295]

Answer:

A) la que cuesta $1.89

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Step-by-step explanation:

5 0
2 years ago
Find the smallest positive $n$ such that \begin{align*} n &\equiv 3 \pmod{4}, \\ n &\equiv 2 \pmod{5}, \\ n &\equiv
Alex777 [14]

4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.

We construct a number x such that taking it mod 4, 5, and 7 leaves the desired remainders:

x=3\cdot5\cdot7+4\cdot2\cdot7+4\cdot5\cdot6

  • Taken mod 4, the last two terms vanish and we have

x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod4

so we multiply the first term by 3.

  • Taken mod 5, the first and last terms vanish and we have

x\equiv4\cdot2\cdot7\equiv51\equiv1\pmod5

so we multiply the second term by 2.

  • Taken mod 7, the first two terms vanish and we have

x\equiv4\cdot5\cdot6\equiv120\equiv1\pmod7

so we multiply the last term by 7.

Now,

x=3^2\cdot5\cdot7+4\cdot2^2\cdot7+4\cdot5\cdot6^2=1147

By the CRT, the system of congruences has a general solution

n\equiv1147\pmod{4\cdot5\cdot7}\implies\boxed{n\equiv27\pmod{140}}

or all integers 27+140k, k\in\mathbb Z, the least (and positive) of which is 27.

3 0
3 years ago
For what interval is the function f(x) = (20 + \sqrt{x}) / (\sqrt{20 + x}) continuous?
konstantin123 [22]
Try this solution:
1. according to the condition
\left \{ {{x \geq 0} \atop {20+x\ \textgreater \ 0}} \right. \ =\ \textgreater \  \ x \geq 0.
2. for more details see the attached graph.

Answer: [0;+oo)

5 0
2 years ago
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