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Artyom0805 [142]
3 years ago
14

Consider the function f(x)=2x√3. For what value of x does f(x)=10?

Mathematics
1 answer:
REY [17]3 years ago
7 0

Answer: 34.64

Step-by-step explanation: you multiply 10 and 2 and that gives you 20 and then u square root 20 and 3 and you get your answer.

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20% de 75 es 50...<span>Es el iniciador</span>
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Solve the system by elimination 4x-y=2 and x+3y=7
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x=1 y=2

Step-by-step explanation:

Isolate x for 4x -y=2: X=2+y/4

Substitute x= 2+y/4

(2+y/4 )+3y=7

simplify 2+13y/4=7

Isolate y for 2+13y/4=7

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Simplify

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So, you go to a Coffee Shop. They have either small, medium or large coffee. They have vanilla or hazelnut flavor. What are the
Talja [164]

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6

Step-by-step explanation:

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6 0
3 years ago
Write the polynomial in factored form as a product of linear factors f(r)=r^3-9r^2+17r-9
adelina 88 [10]

Answer:

  f(r) = (x -1)(x -4+√7)(x -4-√7)

Step-by-step explanation:

The signs of the terms are + - + -. There are 3 changes in sign, so Descartes' rule of signs tells you there are 3 or 1 positive real roots.

The rational roots, if any, will be factors of 9, the constant term. The sum of coefficients is 1 -9 +17 -9 = 0, so you know that r=1 is one solution to f(r) = 0. That means (r -1) is a factor of the function.

Using polynomial long division, synthetic division (2nd attachment), or other means, you can find the remaining quadratic factor to be r^2 -8r +9. The roots of this can be found by various means, including completing the square:

  r^2 -8r +9 = (r^2 -8r +16) +9 -16 = (r -4)^2 -7

This is zero when ...

  (r -4)^2 = 7

  r -4 = ±√7

  r = 4±√7

Now, we know the zeros are {1, 4+√7, 4-√7), so we can write the linear factorization as ...

  f(r) = (r -1)(r -4 -√7)(r -4 +√7)

_____

<em>Comment on the graph</em>

I like to find the roots of higher-degree polynomials using a graphing calculator. The red curve is the cubic. Its only rational root is r=1. By dividing the function by the known factor, we have a quadratic. The graphing calculator shows its vertex, so we know immediately what the vertex form of the quadratic factor is. The linear factors are easily found from that, as we show above. (This is the "other means" we used to find the quadratic roots.)

7 0
3 years ago
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