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3 years ago

Find the other endpoint of the line segment with the given endpoint and midpoint.

Mathematics

1 answer:

OLEGan [10]3 years ago

7 0

Answer:

Other Endpoint (x₂,y₂) = (16,25)

Step-by-step explanation:

We are given endpoint = (4,-9) and midpoint(10,8) we need to find the other endpoint of line segment.

Let other endpoint be (x₂,y₂)

The formula used to find other endpoint is using formula of midpoint

$M(x_m,y_m)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

We have $x_1=4 , y_1= -9 , x_m=10, y_m=8$

Using formula and finding (x₂,y₂)

$M(x_m,y_m)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\\(10,8)=(\frac{4+x_2}{2},\frac{-9+y_2}{2})\\10=\frac{4+x_2}{2} , 8=\frac{-9+y_2}{2}\\Finding\ x_2\\10*2=4+x_2 \\20=4+x_2\\x_2=20-4, \\x_2=16\\Finding \ y_2\\8*2=-9+y_2\\16=-9+y_2\\y_2=16+9\\y_2=25$

So, x₂=16 and y₂=25.

So, Other Endpoint (x₂,y₂) = (16,25)

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A bathtub is filled with 38 1/3 gallons of water. If 2 3/5 gallons splash out, how much water is left in the tube

musickatia [10]

So let's convert these to improper fractions:

$38\frac{1}{3}=\frac{115}{3}$

$2\frac{3}{5}=\frac{13}{5}$

Then let's convert these fractions so that they have a denominator of 15 so we can better compare these numbers:

$\frac{115}{3}=\frac{575}{15}$

$\frac{13}{5}=\frac{39}{15}$

So when we have $\frac{39}{15}$ gallons splash out, we are left with $\frac{536}{15}$ gallons or $35\frac{11}{15}$ gallons in the tub.

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3 years ago

I am lost can anyone help

Rina8888 [55]

I would recommend taking a picture of the instructions because I am not sure if you need to find the equation of the line that is parallel or perpendicular, so I will do both.

1. (3 , 2); y = 3x - 2

If the line is parallel to the given equation, the slopes have to be the SAME, so the slope(m) is 3

y = mx + b

y = 3x + b

To find b you plug in the point (3, 2) into the equation

2 = 3(3) + b

2 = 9 + b

-7 = b

The equation of the line that is parallel to the given equation is:

y = 3x - 7

To find the equation of the line that is perpendicular to the given equation, the slope has to be the exact opposite of the given slope. (you flip the sign and the number of the given slope to get the perpendicular line's slope)

The given slope is 3, the perpendicular slope is $-\frac{1}{3}$

y = mx + b

$y = -\frac{1}{3} x + b$

To find b, you plug in the point (3 , 2) into the equation

$2 = -\frac{1}{3} (3) + b$

2 = -1 + b

3 = b

The equation of the line perpendicular to the given equation is:

$y = -\frac{1}{3} x +3$

8 0

3 years ago

Answer the following question about the function whose derivative is given below

Komok [63]

Answer:

a) The critical points are $x = 3$ and $x = -6$ .

b) f is decreasing in the interval $(-\infty, -6)$

f is increasing in the intervals $(-6,3)$ and $(3,\infty)$ .

c) Local minima: $x = -6$

Local maxima: No local maxima

Step-by-step explanation:

(a) what are the critical points of f?

The critical points of f are those in which $f^{\prime}(x) = 0$ . So

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

$(x-3)^{2}(x+6) = 0$

So, the critical points are $x = 3$ and $x = -6$ .

(b) on what intervals is f increasing or decreasing? (if there is no interval put no interval)

For any interval, if $f^{\prime}$ is positive, f is increasing in the interval. If it is negative, f is decreasing in the interval.

Our critical points are $x = 3$ and $x = -6$ . So we have those following intervals:

$(-\infty, -6), (-6,3), (3, \infty)$

We select a point x in each interval, and calculate $f^{\prime}(x)$ .

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

$(-\infty, -6)$

$f^{\prime}(-7) = (-7-3)^{2}(-7+6) = (100)(-1) = -100$

f is decreasing in the interval $(-\infty, -6)$

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

$(-6,3)$

$f^{\prime}(2) = (2-3)^{2}(2+6) = (1)(8) = 8$

f is increasing in the interval $(-6,3)$ .

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

$(3, \infty)$

$f^{\prime}(4) = (4-3)^{2}(4+6) = (1)(10) = 10$

f is increasing in the interval $(3,\infty)$ .

(c) At what points, if any, does f assume local maximum and minima values. ( if there is no local maxima put mo local maxima) if there is no local minima put no local minima

At a critical point x, if the function goes from decreasing to increasing, it is a local minima. And if the function goes from increasing to decreasing, it is a local maxima.

So, for each critical point is this problem:

At $x = -6$ , f goes from decreasing to increasing.

So $x = -6$ , f assume a local minima value

At $x = 3$ , f goes from increasing to increasing. So, there it is not a local maxima nor a local minima. So, there is no local maxima for this function.

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3 years ago

What is 5y-4(-4-7y) using Distributive property

AlexFokin [52]

Answer:

33y+16

Step-by-step explanation:

Break apart the problem:

-4(-4-7y)=16+28y

5y+16+28y=33y+16 (Add like terms)

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3 years ago

Which shows two expressions that are equivalent to (Negative 7) (Negative 15) (Negative 5) Yan is climbing down a ladder. Each t

Svet_ta [14]

Answer:

(-8) + (-8) + (-8) + (-8)

Step-by-step explanation:

Since Yan is descending ; we represent his movement as being negative.

Number of rungs descended before stopping = 4

Number of stops = 8

Given that no extra steps was taken ;

Number of ladder rungs descended :

4 rungs per stop for 8 stops

- 4 * 8 = - 32 rungs

Another way to write the product :

Number of stops added in number of rungs descended :

(-8) + (-8) + (-8) + (-8)

3 0

2 years ago