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Zigmanuir [339]
2 years ago
8

How to solve the following system y=(1/2)x^2+2x-1 and 3x-y=1

Mathematics
1 answer:
ruslelena [56]2 years ago
7 0

Answer:

(0,-1),(2,5)

Step-by-step explanation:

3x-y=1

0.5x^{2}+2x-1=y

Substitute the equation of y into the other equation

3x-(0.5x^{2}+2x-1)=1

3x-0.5x^{2}-2x+1=1

collect like terms

-0.5x^{2}+x=0

Solve for x

x(-0.5x+1)=0

0=-0.5x+1

-1=-0.5x

x=2 x=0

Back substitute into the other original equation and solve for y

3(2)-y=1

6-y=1

y=5

(2,5)

3(0)-y=1

-1(-y)=-1(1)

y=-1

(0,-1)

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Anvisha [2.4K]

Answer:

Step-by-step explanation:

We'll take this step by step.  The equation is

8-3\sqrt[5]{x^3}=-7

Looks like a hard mess to solve but it's actually quite simple, just do one thing at a time.  First thing is to subtract 8 from both sides:

-3\sqrt[5]{x^3}=-15

The goal is to isolate the term with the x in it, so that means that the -3 has to go.  Divide it away on both sides:

\sqrt[5]{x^3}=5

Let's rewrite that radical into exponential form:

x^{\frac{3}{5}}=5

If we are going to solve for x, we need to multiply both sides by the reciprocal of the power:

(x^{\frac{3}{5}})^{\frac{5}{3}}=5^{\frac{5}{3}}

On the left, multiplying the rational exponent by its reciprocal gets rid of the power completely.  On the right, let's rewrite that back in radical form to solve it easier:

x=\sqrt[3]{5^5}

Let's group that radicad into groups of 3's now to make the simplifying easier:

x=\sqrt[3]{5^3*5^2} because the cubed root of 5 cubed is just 5, so we can pull it out, leaving us with:

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8 0
3 years ago
Describe how the range of a data set can help describe its variability.
katovenus [111]

Answer:

The range is defined as the difference between the term with the highest value and the term with the lowest value. This statistic is used to measure the variability of a series of data because it provides information on how far apart the values of a tail of the distribution are from the values at the other end of the tail.

Imagine that you manufacture a type of spare part for cars that must have a measurement of 10 cm with a margin of error of 1 cm.

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10 ± 1 cm

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If you randomly select a sample of n pieces and measure them, the variability is expected to be low, so that your process is of quality, then expect a low range preferably less than 1 cm.

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But if you find that the range is up to 8 cm, this would mean that not all pieces measure around 10 cm, it means that the variability of the measurements is high.

{14, 12, 11, 8, 7, 11, 12, 15} Range = 15 - 7 = 8 cm   <em>high variability </em>

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