Review questions

Write an inequality to represent the following scenario: Hagrid’s goal at the National Hotdog Eating Contest is to consume 32 ho

kotykmax [81]

To find the value of a variance add and subtract the variance from the original number.

The original number is 32 hot dogs.

The variance is 5 hot dogs.

32-5 = 27

32 + 5 = 37

This means he needs to eat between 27 and 37 hot dogs.

he inequality with variance is written as: 27 < h < 37

6 0

3 years ago

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A computer technician charges $60 to come to a home to repair a

-Dominant- [34]

5 hours
Hope this helps!

4 0

3 years ago

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The remainder after diving x^4+3x^3-8x^2+5x-9 by x+5 is ____.

raketka [301]

Use the polynomial remainder theorem. It says that a polynomial $f(x)$ has remainder $f(c)$ upon division by $x-c$ .

Here we have

$f(x)=x^4+3x^3-8x^2+5x-9$

and $c=-5$ , so the remainder is

$f(-5)=(-5)^4+3(-5)^3-8(-5)^2+5(-5)-9=\boxed{16}$

6 0

3 years ago

A candle in the shape of a circular cone has a base of radius r and a height of h that is the same length as the radius. which e

ladessa [460]

Answer:

$\frac{r(1-\sqrt{2})}{-3}$

Step-by-step explanation:

Volume of cone = $\frac{1}{3} \pi r^{2} h$

Since we are given that a circular cone has a base of radius r and a height of h that is the same length as the radius

= $\frac{1}{3} \pi r^{2} \times r$

= $\frac{1}{3} \pi r^{3}$

Surface area of cone including 1 base = $\pi r^{2} +\pi\times r \times \sqrt{r^2+h^2}$

Since r = h

So, area = $\pi r^{2} +\pi\times r \times \sqrt{r^2+r^2}$

= $\pi r^{2} +\pi\times r \times \sqrt{2r^2}$

= $\pi r^{2} +\pi\times r^2 \times \sqrt{2}$

Ratio of volume of cone to its surface area including base :

$\frac{\frac{1}{3} \pi r^{3}}{\pi r^{2} +\pi\times r^2 \times \sqrt{2}}$

$\frac{\frac{1}{3}r}{1+\sqrt{2}}$

$\frac{r}{3(1+\sqrt{2})}$

Rationalizing

$\frac{r}{3(1+\sqrt{2})} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}$

$\frac{r(1-\sqrt{2})}{-3}$

Hence the ratio the ratio of the volume of the candle to its surface area(including the base) is $\frac{r(1-\sqrt{2})}{-3}$

8 0

3 years ago

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Without plotting points, let M=(-2,-1), N=(3,1), M'= (0,2), and N'=(5, 4). Without using the distanceformula, show that segments

kramer

Given:

M=(x1, y1)=(-2,-1),

N=(x2, y2)=(3,1),

M'=(x3, y3)= (0,2),

N'=(x4, y4)=(5, 4).

We can prove MN and M'N' have the same length by proving that the points form the vertices of a parallelogram.

For a parallelogram, opposite sides are equal

If we prove that the quadrilateral MNN'M' forms a parallellogram, then MN and M'N' will be the oppposite sides. So, we can prove that MN=M'N'.

To prove MNN'M' is a parallelogram, we have to first prove that two pairs of opposite sides are parallel,

Slope of MN= Slope of M'N'.

Slope of MM'=NN'.

$\begin{gathered} \text{Slope of MN=}\frac{y2-y1}{x2-x1} \\ =\frac{1-(-1)}{3-(-2)} \\ =\frac{2}{5} \\ \text{Slope of M'N'=}\frac{y4-y3}{x4-x3} \\ =\frac{4-2}{5-0} \\ =\frac{2}{5} \end{gathered}$

Hence, slope of MN=Slope of M'N' and therefore, MN parallel to M'N'

$\begin{gathered} \text{Slope of MM'=}\frac{y3-y1}{x3-x1} \\ =\frac{4-(-1)}{5-(-2)} \\ =\frac{3}{2} \\ \text{Slope of NN'=}\frac{y4-y2}{x4-x2} \\ =\frac{4-1}{5-3} \\ =\frac{3}{2} \end{gathered}$

Hence, slope of MM'=Slope of NN' nd therefore, MM' parallel to NN'.

Since both pairs of opposite sides of MNN'M' are parallel, MM'N'N is a parallelogram.

Since the opposite sides are of equal length in a parallelogram, it is proved that segments MN and M'N' have the same length.

7 0

1 year ago