What is the range of the function y = x ^2?

Can someone anyone please man help me out

zzz [600]

Answer:

2 sqrt(3) -2i

Step-by-step explanation:

If we have the magnitude and the theta

z = |z| cos theta + i |z| sin theta

= 4 cos 330 + i 4 sin 330

=4 (sqrt(3)/2) + 4i (-1/2)

= 2 sqrt(3) -2i

5 0

3 years ago

Please answer. find volume. thanks!

aivan3 [116]

Answer:

VOLUME = L X W X H AND THAT EQUALS TO 250 1/2

Step-by-step explanation:

3 0

1 year ago

The linear function f(x) = 0.9x + 79 represents the average test score in your math class, where x is the number of the test tak

Levart [38]

Given:

The linear function for average test score in your math class is

$f(x)=0.9x+79$

where x is the number of the test taken.

The linear function g(x) in the given table represents the average test score in your science class, where x is the number of the test taken.

To find:

Part A: The test average for your math class after completing test 2.

Part B: The test average for your science class after completing test 2.

Part C: Which class had a higher average after completing test 4?

Solution:

Part A:

We have,

$f(x)=0.9x+79$

Put x=2 in the above function, to find the test average for your math class after completing test 2.

$f(2)=0.9(2)+79$

$f(2)=1.8+79$

$f(2)=80.8$

Therefore, the test average for your math class after completing test 2 is 80.8.

Part B:

From the given table it is clear that g(x) =79 at x=2.

Therefore, the test average for your science class after completing test 2 is 79.

Part C:

Put x=4 in f(x), to find the test average for your math class after completing test 4.

$f(4)=0.9(4)+79$

$f(4)=3.6+79$

$f(4)=82.6$

From the table it is clear that the value of g(x) is increased by 1 as x is increased by 1. So, the value of function g(x) at x=4 must be 1 more than 80.

$g(4)=81$

It is clear that,

$82.6>81$

Therefore, math class had a higher average after completing test 4.

3 0

3 years ago

Arnie decides to make 3 cups of dark orange paint and 3 cups of light orange paint.

dimulka [17.4K]

The question is an illustration of equivalent ratios.

3 cups of dark orange require 3 ounces of yellow paint
3 cups of light orange require 19.5 ounces of yellow paint

1 cup of dark orange requires

$Red = 7oz$

$Yellow =1oz$

To make 3 cups, we simply multiply by 3, because $3 \times 1 = 3$

So, we have:

$Red = 7oz \times 3 = 21oz$

$Yellow =1oz \times 3 = 3oz$

2 cups of light orange require

$Red =3oz$

$Yellow =13$

To make 3 cups, we simply multiply by 1.5, because $2 \times 1.5 = 3$

So, we have:

$Red = 3oz \times 1.5 = 4.5oz$

$Yellow = 13oz \times 1.5 = 19.5oz$

Hence:

3 cups of dark orange require 3 ounces of yellow paint
3 cups of light orange require 19.5 ounces of yellow paint

Read more about equivalent ratios at:

brainly.com/question/18441891

5 0

2 years ago

Resistors are labeled 100 Ω. In fact, the actual resistances are uniformly distributed on the interval (95, 103). Find the mean

Zinaida [17]

Answer:

$E[R]$ = 99 Ω

$\sigma_R$ = 2.3094 Ω

P(98<R<102) = 0.5696

Step-by-step explanation:

The mean resistance is the average of edge values of interval.

Hence,

The mean resistance, $E[R] = \frac{a+b}{2} = \frac{95+103}{2} = \frac{198}{2}$ = 99 Ω

To find the standard deviation of resistance, we need to find variance first.

$V(R) = \frac{(b-a)^2}{12} =\frac{(103-95)^2}{12} = 5.333$

Hence,

The standard deviation of resistance, $\sigma_R = \sqrt{V(R)} = \sqrt5.333$ = 2.3094 Ω

To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.

$z_1 = \frac{102-99}{2.3094} = 1.299$

$z_2 = \frac{98-99}{2.3094} = -0.433$

From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696

5 0

3 years ago