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3 years ago

M is the midpoint of segment GR. G has coordinates (11,-5) and M is (4,-4) Find the coordinates of R

Mathematics

1 answer:

7nadin3 [17]3 years ago

4 0

Answer:

(3, -3)

Step-by-step explanation:

Given:

G = (11,-5)
M = (4,-4)
R = ?

Solution

R(x, y)

As per midpoint formula;

4 = (11 + x)/2 ⇒ x + 11 = 8 ⇒ x = 3
-4 = (-5 + y)/2 ⇒ y - 5 = -8 ⇒ y = -3
R = (3, -3)

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What is the solution to this inequality: -4(3x-5)<2(3x+4)

Delicious77 [7]

Answer:

x > 2/3

Step-by-step explanation:

-4(3x - 5) < 2(3x + 4)

Divide both sides by 2.

-2(3x - 5) < 3x + 4

Distribute on the left side.

-6x + 10 < 3x + 4

Subtract 3x from both sides.

-9x + 10 < 4

Subtract 10 from both sides.

-9x < - 6

Divide both sides by -9. Remember that when you divide both sides of an inequality by a negative number, the inequality sign changes direction.

x > (-6)/(-9)

x > 6/9

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3 years ago

Hello people ~

Luden [163]

Cone details:

height: h cm

radius: r cm

Sphere details:

radius: 10 cm

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

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

Using Pythagoras Theorem

(a)

TO² + TU² = OU²

(h-10)² + r² = 10² [insert values]

r² = 10² - (h-10)² [change sides]

r² = 100 - (h² -20h + 100) [expand]

r² = 100 - h² + 20h -100 [simplify]

r² = 20h - h² [shown]

r = √20h - h² ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

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

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (\sqrt{20h - h^2})^2 \ ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20h - h^2) (h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20 - h) (h) ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)$

To find maximum/minimum, we have to find first derivative.

(c)

First derivative

$\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )$

apply chain rule

$\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}$

Equate the first derivative to zero, that is V'(x) = 0

$\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0$

$\Longrightarrow \sf 40h-3h^2=0$

$\Longrightarrow \sf h(40-3h)=0$

$\Longrightarrow \sf h=0, \ 40-3h=0$

$\Longrightarrow \sf h=0,\:h=\dfrac{40}{3}$

maximum volume: when h = 40/3

$\sf \Longrightarrow max= \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )$

$\sf \Longrightarrow maximum= 1241.123 \ cm^3$

minimum volume: when h = 0

$\sf \Longrightarrow min= \dfrac{1}{3} \pi (0)^2(20-0)$

$\sf \Longrightarrow minimum=0 \ cm^3$

6 0

2 years ago

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F(-7) = -3

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SO A

8 0

3 years ago

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Novosadov [1.4K]

Answer:

Step-by-step explanation:

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3 years ago

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mars1129 [50]

Answer:

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Step-by-step explanation:

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120 pine trees are going to be planted in the park.

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