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2 years ago

7

For the line and point on the picture, find a perpendicular line, through the given point. please look at the picture to see the

point and line and answer choices.

1 answer:

miv72 [106K]2 years ago

5 0

Answer:

y = - 7

Step-by-step explanation:

You might be interested in

Whats the value of x

fiasKO [112]

Answer:

x=14

Step-by-step explanation:

you should set up your problem like this:

5x=3x+28

use order of operations aka PEMDAS to solve

you're answer is 14

8 0

2 years ago

Can someone help me with 3 a and b?? Thank you

polet [3.4K]

M∠ADB + m∠BDC = m∠ADC [<span>angle addition postulate]

x + x + 10 = 60
2x + 10 = 60
2x = 60 - 10
2x = 50
x = 50/2
x = 25

a
m</span>∠ADB = x = 25°

b
m∠BDC = x + 10 = 25 + 10 = 35°

5 0

3 years ago

A hovercraft takes off from a platform. It’s height (in meters), x seconds is modeled by h(x)=-(x-11)(x+3). What is the height o

Eddi Din [679]

Answer:

The height of the hovercraft at the time of takeoff is 33 meters.

Step-by-step explanation:

Height of hovercraft x seconds after takeoff = h(x) = -(x - 11)(x + 3)

Height after takeoff is = h(0) = -(0 - 11)(0 + 3) = -(-11)(3) = 33

So the correct answer is 33 meters. The height of the hover craft at the time of the takeoff is 33 metres.

Answer: 33 meters.

<em><u></u></em>

<em><u>Please mark brainliest</u></em>

<em><u>Hope this helps</u></em>

6 0

3 years ago

Answer it answer it answer it

AysviL [449]

Answer: D.

Step-by-step explanation: If you look closely, putting a 100 wouldn't make sense

6 0

3 years ago

Read 2 more answers

A ball is thrown into the air from a height of 4 feet at time t = 0. The function that models this situation is h(t) = -16t2 + 6

katrin2010 [14]

Answer:

Part a) The height of the ball after 3 seconds is $49\ ft$

Part b) The maximum height is 66 ft

Part c) The ball hit the ground for t=4 sec

Part d) The domain of the function that makes sense is the interval

[0,4]

Step-by-step explanation:

we have

$h(t)=-16t^{2} +63t+4$

Part a) What is the height of the ball after 3 seconds?

For t=3 sec

Substitute in the function and solve for h

$h(3)=-16(3)^{2} +63(3)+4=49\ ft$

Part b) What is the maximum height of the ball? Round to the nearest foot.

we know that

The maximum height of the ball is the vertex of the quadratic equation

so

Convert the function into a vertex form

$h(t)=-16t^{2} +63t+4$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$h(t)-4=-16t^{2} +63t$

Factor the leading coefficient

$h(t)-4=-16(t^{2} -(63/16)t)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$h(t)-4-16(63/32)^{2}=-16(t^{2} -(63/16)t+(63/32)^{2})$

$h(t)-(67,600/1,024)=-16(t^{2} -(63/16)t+(63/32)^{2})$

Rewrite as perfect squares

$h(t)-(67,600/1,024)=-16(t-(63/32))^{2}$

$h(t)=-16(t-(63/32))^{2}+(67,600/1,024)$

the vertex is the point (1.97,66.02)

therefore

The maximum height is 66 ft

Part c) When will the ball hit the ground?

we know that

The ball hit the ground when h(t)=0 (the x-intercepts of the function)

so

$h(t)=-16t^{2} +63t+4$

For h(t)=0

$0=-16t^{2} +63t+4$

using a graphing tool

The solution is t=4 sec

see the attached figure

Part d) What domain makes sense for the function?

The domain of the function that makes sense is the interval

[0,4]

All real numbers greater than or equal to 0 seconds and less than or equal to 4 seconds

Remember that the time can not be a negative number

6 0

3 years ago