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3 years ago

Graph y = −x^2 − 1. Identify the vertex of the graph. Tell whether it is a minimum or maximum.

Mathematics

1 answer:

bonufazy [111]3 years ago

5 0

Answer:

Step-by-step explanation:

since this is an upside down parabola shifted down 1 unit, the vertex must be at the point (0,-1). It must be a <em>MAX</em> because the parabola is upside down

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Can someone help me please I need the mode, median, range and the mean.?

Neko [114]

Answer:

First the mode. Since 5 popped up the most, 5 is the mode.

Next is the median. I crossed 1 dot from each side until it shows the last dot, and 5 was the last one.

After that the range. 9-2=7

Finally the worst, the mean... 2+2+3+3+3+4+5+5+5+5+5+6+6+6+8+9+9+9+9

=104/19=5.47

SO, Mode=5 Median=5 Range=7, and the mean is 5.47 (rounded nearest hundred)

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3 years ago

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EleoNora [17]

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2 years ago

How to find imaginary zeros anf real zeros of F(x)=-4x^5-8x^3+12x

Fofino [41]

Answer:

$x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}$

Step-by-step explanation:

We are given the function:

$f(x)=-4x^5-8x^3+12x$

And we want to finds its zeros.

Therefore:

$0=-4x^5-8x^3+12x$

Firstly, we can divide everything by -4:

$0=x^5+2x^3-3x$

Factor out an x:

$0=x(x^4+2x^2-3)$

This is in quadratic form. For simplicity, we can let:

$u=x^2$

Then by substitution:

$0=x(u^2+2u-3)$

Factor:

$0=x(u+3)(u-1)$

Substitute back:

$0=x(x^2+3)(x^2-1)$

By the Zero Product Property:

$x=0\text{ and } x^2+3=0\text{ and } x^2-1=0$

Solving for each case:

$x=0\text{ and } x=\pm\sqrt{-3}\text{ and } x=\pm\sqrt{1}$

Therefore, our real and complex zeros are:

$x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}$

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2 years ago

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Alecsey [184]

Answer:

the correct answer is 16

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3 years ago

Find the unit price for each size of Jiffy peanut butter and determine which size is the better buy 18 oz. $3.28

Liono4ka [1.6K]

Price this per ounce
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3 years ago