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3 years ago

Which ordered pair is the vertex of y=|x-2|+1

Mathematics

1 answer:

swat323 years ago

8 0

Answer:

(-2, -1)

Step-by-step explanation:

-2 is x

-1 is y

thats all i know

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A 12-ft-by-15-ft rectangular swimming pool has a 3-ft-wide no-slip surface around it. What is the outer perimeter of the no slip

Nataliya [291]

Answer: 74 ft
12+12+15+15= 74
Perimeter is the distance around sometime

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3 years ago

A useful rule of thumb called the "rule of 70" states that if something grows at a constant rate of z percent per year, it doubl

vesna_86 [32]

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4 years ago

Find the volume of the solid formed by revolving the region bounded by LaTeX: y = \sqrt{x} y = x and the lines LaTeX: y = 1 y =

Strike441 [17]

Answer:

The volume is:

$\displaystyle\frac{37\pi}{10}$

Step-by-step explanation:

See the sketch of the region in the attached graph.

We set the integral using washer method:

$\displaystyle\int_a^b\pi r^2dx$

Notice here the radius of the washer is the difference of the given curves:

$x-\sqrt{x}$

So the integral becomes:

$\displaystyle\int_1^4\pi(x-\sqrt{x})^2dx$

We solve it:

Factor $\pi$ out and distribute the exponent (you can use FOIL):

$\displaystyle\pi\int_1^4x^2-2x\sqrt{x}+x\,dx$

Notice: $x\sqrt{x}=x\cdot x^{1/2}=x^{3/2}$

So the integral becomes:

$\displaystyle\pi\int_1^4x^2-2x^{3/2}+x\,dx$

Then using the basic rule to evaluate the integral:

$\displaystyle\pi\left[\frac{x^3}{3}-\frac{2x^{5/2}}{5/2}+\frac{x^2}{2}\right|_1^4$

Simplifying a bit:

$\displaystyle\pi\left[\frac{x^3}{3}-\frac{4x^{5/2}}{5}+\frac{x^2}{2}\right|_1^4$

Then plugging the limits of the integral:

$\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(4)^{5/2}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right]$

Taking the root (rational exponents):

$\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(2)^{5}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right]$

Then doing those arithmetic computations we get:

$\displaystyle\frac{37\pi}{10}$

6 0

3 years ago

Find [5(cos 330 degrees + I sin 330 degrees)]^3

earnstyle [38]

Given a complex number in the form:
$z= \rho [\cos \theta + i \sin \theta]$
The nth-power of this number, $z^n$ , can be calculated as follows:

- the modulus of $z^n$ is equal to the nth-power of the modulus of z, while the angle of $z^n$ is equal to n multiplied the angle of z, so:
$z^n = \rho^n [\cos n\theta + i \sin n\theta ]$
In our case, n=3, so $z^3$ is equal to
$z^3 = \rho^3 [\cos 3 \theta + i \sin 3 \theta ] = (5^3) [\cos (3 \cdot 330^{\circ}) + i \sin (3 \cdot 330^{\circ}) ]$ (1)
And since
$3 \cdot 330^{\circ} = 990^{\circ} = 2\pi +270^{\circ}$
and both sine and cosine are periodic in $2 \pi$ , (1) becomes
$z^3 = 125 [\cos 270^{\circ} + i \sin 270^{\circ} ]$

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4 years ago

How many edges does a square based pyramid have

Vikki [24]

Answer:

Step-by-step explanation:

5 0

4 years ago