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3 years ago

Which system of equations can be used to find the roots of the equation 4x^2=x^3+2x?

Mathematics

2 answers:

kow [346]3 years ago

6 0

The given equation is 4x^2=x^3+2x.

On the left side it has 4x² and on the right side the expression is x³+2x.

So, if we have any system of equations in which we have these two expression which will be equal to any other variable then we can use that system of equations to find the roots.

Now notice that in option D it's given,

y=4x² and y=x³+2x.

So, if we will equate this equations then we will get the same above equation which will be use to find the roots.

So, D is the correct choice.

salantis [7]3 years ago

6 0

The answer is D on edge

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If alpha and beta are the zeroes of the polynomial 6x2+x-2 find the value og alpha/beta + beta/alpha

castortr0y [4]

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$use\ Vieta's\ formula:\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{\alpha^2+2\alpha\beta+\beta^2-2\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2}{\alpha\beta}-2\\\\\alpha+\beta=\frac{-b}{a};\ \alpha\beta=\frac{c}{a}\\\\\frac{(\alpha+\beta)^2}{\alpha\beta}-2=\frac{\left(\frac{-b}{a}\right)^2}{\frac{c}{a}}-2=\frac{b^2}{a^2}\cdot\frac{a}{c}-2=\frac{b^2}{ac}-2$

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3 years ago

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Answer:

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3 years ago

The proportion of households in a region that do some or all of their banking on the Internet is 0.31. In a random sample of 100

Alenkasestr [34]

Answer:

Approximate probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is less than 0.0005% .

Step-by-step explanation:

We are given that let X be the number that do some or all of their banking on the Internet.

Also; Mean, $\mu$ = 310/1000 or 0.31 and Standard deviation, $\sigma$ = 14.63/1000 = 0.01463 .

We know that Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

Probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is given by P(X <= 130/1000);

P(X <=0.13) = P( $\frac{X-\mu}{\sigma}$ <= $\frac{0.13-0.31}{0.01463}$ ) = P(Z <= -12.303) = P(Z > 12.303)

Since this value is not represented in the z table as the value is very high and z table is limited to x = 4.4172.

So, after seeing the table we can say that this probability is approximately less than 0.0005% .

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