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3 years ago

What is the distance between the points (5, −2) and (−9, −2)?

Mathematics

2 answers:

Dima020 [189]3 years ago

6 0

Answer:

D.14

Step-by-step explanation:

-9 is 14 below 5

Sergeeva-Olga [200]3 years ago

4 0

Answer:

The answer would be D.

Step-by-step explanation:

Use the website math-way without that > -

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Round 219.870754015 to the nearest hundred.

KatRina [158]

Answer:

200

Step-by-step explanation:

Because you are rounding to the hundreds, you look at the tens place. The tens place is a "1" and 1 is below five, so you round down to 200.

4 0

3 years ago

Read 2 more answers

Read this and answer this for me Thanks <3

Norma-Jean [14]

Answer:

The answer is 1st point

Step-by-step explanation:

(12x²+4x-5)-(9x²-9x+1)

= 12x²+4x-5-9x²+9x-1

= 3x²+13x-6

6 0

3 years ago

Consider the following function. f(x) = 7x + 1 x (a) Find the critical numbers of f. (Enter your answers as a comma-separated li

sukhopar [10]

Answer:

a) DNE

b) The function increases for every real value of x.

c) DNE

Step-by-step explanation:

Given a function f(x), the critical points are those in which $f^{\prime}(x) = 0$ , that is, the roots of the first derivative of f(x).

Those critical points let us find the intervals in which the function increases or decreases. If the first derivative in the interval is positive, the function increases in the interval. If it is negative, the function decreases.

If the function increases before a critical point and then, as it passes the critical point, it starts to decrease, we have that the critical point $(x_{c}, f(x_{c}) is a relative maximum.If the function decreases before a critical point and then, as it passes the critical point, it starts to increase, we have that the critical point [tex](x_{c}, f(x_{c}) is a relative minimum.If the function has no critical points, it either always increases or always decreases.In this exercise, we have that:[tex]f(x) = 7x + 1$

(a) Find the critical numbers of f.

$f^{\prime}(x) = 0$

$f^{/prime}(x) = 7$

7 = 0 is false. This means that f has no critical points.

(b) Find the open intervals on which the function is increasing or decreasing.

Since there are no critical points, we know that either the function increases or decreases in the entire real interval.

We have a first order function in the following format:

$f(x) = ax + b$

In which $a > 0$ .

So the function increases for every real value of x.

(c) Apply the First Derivative Test to identify the relative extremum.

From a), we find that there are no critical numbers. So DNE

7 0

3 years ago

Owen simplified the expression r-8 s-5.<br><br><br>r−8s−5 = 1r8 ⋅ s5 = s5r8

kobusy [5.1K]

r⁻⁸ s⁻⁵

using the rule x⁻ⁿ=1/xⁿ

r⁻⁸ s⁻⁵ = (1/r⁸)(1/s⁵)

6 0

3 years ago

Read 2 more answers

21. Who is closer to Cameron? Explain.

pickupchik [31]

Problem 21

<h3>Answer: Jamie is closer</h3>

-----------------------

Explanation:

A = Arthur's location = (20,35)
J = Jamie's location = (45,20)
C = Cameron's location = (65,40)

To find out who's closer to Cameron, we need to compute the segment lengths AC and JC. Then we pick the smaller of the two lengths.

We use the distance formula to find each length

Let's find the length of AC.

$A = (x_1,y_1) = (20,35)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from A to C} = \text{length of segment AC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-65)^2 + (35-40)^2}\\\\d = \sqrt{(-45)^2 + (-5)^2}\\\\d = \sqrt{2025 + 25}\\\\d = \sqrt{2050}\\\\d \approx 45.2769257\\\\$

The distance from Arthur to Cameron is roughly 45.2769257 units.

Let's repeat this process to find the length of segment JC

$J = (x_1,y_1) = (45,20)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from J to C} = \text{length of segment JC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(45-65)^2 + (20-40)^2}\\\\d = \sqrt{(-20)^2 + (-20)^2}\\\\d = \sqrt{400 + 400}\\\\d = \sqrt{800}\\\\d \approx 28.2842712\\\\$

Going from Jamie to Cameron is roughly 28.2842712 units

We see that segment JC is shorter than AC. Therefore, Jamie is closer to Cameron.

=================================================

Problem 22

<h3>Answer: Arthur is closest to the ball</h3>

-----------------------

Explanation:

We have these key locations:

A = Arthur's location = (20,35)
J = Jamie's location = (45,20)
C = Cameron's location = (65,40)
B = location of the ball = (35,60)

We'll do the same thing as we did in the previous problem. This time we need to compute the following lengths:

These segments represent the distances from a given player to the ball. Like before, the goal is to pick the smallest of these segments to find out who is the closest to the ball.

The steps are lengthy and more or less the same compared to the previous problem (just with different numbers of course). I'll show the steps on how to get the length of segment AB. I'll skip the other set of steps because there's only so much room allowed.

$A = (x_1,y_1) = (20,35)\\\\B = (x_2,y_2) = (35,60)\\\\d = \text{Distance from A to B} = \text{length of segment AB}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-35)^2 + (35-60)^2}\\\\d = \sqrt{(-15)^2 + (-25)^2}\\\\d = \sqrt{225 + 625}\\\\d = \sqrt{850}\\\\d \approx 29.1547595\\\\$

Segment AB is roughly 29.1547595 units.

If you repeated these steps, then you should get these other two approximate segment lengths:

JB = 41.2310563

CB = 36.0555128

-------------

So in summary, we have these approximate segment lengths

AB = 29.1547595
JB = 41.2310563
CB = 36.0555128

Segment AB is the smallest of the trio, which therefore means Arthur is closest to the ball.

3 0

3 years ago