2x + 4 = 3(x – 2) + 1 simplify

What is the domain, range, and function? {(-3, 3), (1, 1), (0, -2), (1,4), (5, -1)}

wolverine [178]

The domain are all the inputs, that is; the x-values

Domain = { -3, 0, 1, 5}

The range are all the output, that is the y-values

Range = { 3, 1, -2, 4, -1}

This is just a relation and not a function, as we have more than 2 same value of x

8 0

1 year ago

PLEASE HELP ME WITH THIS QUESTION!!! WHATS THE ANSWER?? I WILL GIVE YOU BRAINLIEST AND A THANKS!!! THANK YOUUU!!! <33

Free_Kalibri [48]

Answer:

A) 113 m²

Step-by-step explanation:

Use the formula for area of a circle.

Area of a circle = πr²

π = pi = 3.14 (rounded)

r = radius

² = the power of 2, or the radius (in this case) multiplied by itself.

Note that the radius is given, and it is 6. Plug in 6 for the radius:

A (circle) = π(6)²

First, solve the power. Multiply 6²:

6² = 6 * 6 = 36

Next, multiply 36 with π (3.14):

36 x 3.14 = 113.04

A) 113 m² is your closest answer.

~

6 0

2 years ago

Read 2 more answers

The ratio of the areas of two similar rectangles is 1:4. What is the ratio of the corresponding sides?

levacccp [35]

The ratio of linear dimensions (side lengths) is the square root of the ratio of areas.
√1 : √4 = 1 : 2

The best choice is ...
B. 1:2

4 0

2 years ago

What's the mean of 3 4 5 3 4 5 6 7 1 2 3 4 5 6 4 8 4 3 2 A. 4.1 B.4.2 C.8.0 D.8.3

Bess [88]

Answer:

A. 4.1

Step-by-step explanation:

3+4+ 5+ 3+ 4+ 5+ 6+ 7+ 1+ 2+ 3+ 4+ 5+ 6+ 4+ 8+ 4+ 3+ 2= 79

79/ 19 = 4.1

3 0

3 years ago

A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,

algol13

Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}$

Where $v_i$ is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is $[0, \pi/2]$ . The critical points of the function are those who make $d'(\theta)=0$ :

$d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}$

$d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...$

The critical value inside the interval is $\pi/4$ .

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft$

The second step is to find the values of the function at the endpoints of the interval:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft$

The biggest value of f is gived by $\pi/4$ , therefore $\pi/4$ is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

4 0

3 years ago