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3 years ago

What is the midpoint ?

Mathematics

1 answer:

inn [45]3 years ago

8 0

Answer:

Step-by-step explanation:

A(-5,-4) B(-3, 3)

(-5 + -3)/2 = -8/2= -4

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

(-4, -1/2) the midpoint

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Math help pls thank u c:

Verdich [7]

Answer: First one is B second one just make it

Step-by-step explanation:

4 0

3 years ago

Are AB and CD parallel, perpendicular, or neither?*<br> 1. A(-4, 8), B(4, 10), C(1, 1), D(-2, 13)

ohaa [14]

Answer:

Perpendicular

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Step-by-step explanation:

Given

$A = (-4, 8)$

$B = (4, 10)$

$C = (1, 1)$

$D = (-2, 13)$

Required

Is AB and CD parallel?

First, we need to calculate the slope (m) of AB and CD

$m = \frac{y_2 - y_1}{x_2 - x_1}$

For AB:

$A (x_1,y_1)= (-4, 8)$

$B(x_2,y_2) = (4, 10)$

So:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{10 - 8}{4 - (-4)}$

$m = \frac{10 - 8}{4 +4}$

$m = \frac{2}{8}$

$m = \frac{1}{4}$

For CD:

$C(x_1,y_1) = (1, 1)$

$D(x_2,y_2) = (-2, 13)$

So:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{13 - 1}{-2 - 1}$

$m = \frac{12}{-3}$

$m = -4$

For Lines to be parallel, the slope must be equal:

i.e. $m_1 = m_2$

This condition is not true because:

$\frac{1}{4} \neq -4$

For Lines to be perpendicular, the slope must be:

$m_1 = -\frac{1}{m_2}$

This implies:

$\frac{1}{4} = -\frac{1}{-4}$

$\frac{1}{4} = \frac{-1}{-4}$

$\frac{1}{4} = \frac{1}{4}$

<em>Hence, the lines are perpendicular</em>

6 0

3 years ago

Can someone help me please with this!

podryga [215]

Microfilament- they are tiny rods that are comprised of actin filaments.

Herbicide- A substance that is toxic to plants, used to destroy unwanted vegetation.

Arthropod- Any of the large group of animals with jointed limbs and a body made up of segments.

8 0

1 year ago

A cereal box is 12cm tall. The area of the base is 7cm. What is the volume?

andrey2020 [161]

Answer:

588cm3

Step-by-step explanation:

Since the base is 7cm, and it is a cereal box, that means the depth is also 7cm.

Hence,

7x12x7=588

7 0

3 years ago

a projectile is thrown upward so that it's distance above the ground after T seconds is H equals -16t^2 + 672 T. After how many

Arada [10]

Step-by-step explanation:

T = time in seconds

H = distance

Tground = time to return to ground

Tmax = time at maximum height

H = -16T^2 + 672T...eqn 1

projectile returns to ground at H =0,

subs for H in eqn 1...

0 = -16T^2 + 672T

solving for T we get...

16T^2 = 672T

=> Tground = 42secs

Tmax = 0.5 Tground = 21secs

4 0

3 years ago