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3 years ago

7

Please help me with these 4 question i will mark brainliest to correct answer please dont play around

1 answer:

crimeas [40]3 years ago

8 0

Answer:

6, the property of addition.

Step-by-step explanation:

If you add 6 to both sides you used addition lol. Hope I helped, I'm not too good at explaining math sorry

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Brandi, an amateur skier, begins on the "bunny hill," which has a vertical drop of 200 feet. Her speed at the bottom of the hill

Vika [28.1K]

Answer:

Brandi's velocity at the bottom of the hill is 0 ft / s

Step-by-step explanation:

Given:

Height of the bunny hill = 200

speed is given by $v^2 = 64h$

To find:

Brandi's velocity at the bottom of the hill = ?

Solution:

At the bottom of the hill , the height is zero

Thus the speed of Brandi at height h = 0

$v^2 = 64(0)$

$v^2 = 0$

7 0

3 years ago

On Monday, your bank account balance was −$12.83. Because you didn't realize this, you wrote a check for $30.89 for groceries.

Ira Lisetskai [31]

Answer:

You owe $-43.72.

Step-by-step explanation:

Your balance was $-12.83 on Monday. You are writing a check for money you do not have, so you will owe more money.

-$12.83 - $30.89 = -$43.72

4 0

3 years ago

Help please find the graph to the equation

schepotkina [342]

I think it’s option 4. The bottom right one. Hope that’s right!

8 0

3 years ago

Read 2 more answers

Please solve these equations much appreciated

Reptile [31]

Answer:

Part c) $V=16,500\ m^3$

Part d) $V=93,312\pi\ mm^3$ or $V=292,999.68\ mm^3$

Step-by-step explanation:

Part c) Find the volume of the inverted rectangular pyramid

The volume of the rectangular pyramid is equal to

$V=\frac{1}{3}Bh$

where

B is the area of the rectangular base

h is the height of pyramid

<em>Find the area of the base B</em>

$B=(60)(15)=900\ m^2$

$h=55\ m$

substitute

$V=\frac{1}{3}(900)(55)$

$V=16,500\ m^3$

Part d) Find the maximum volume of tea in the tea cup

we know that

The volume of cylinder (tea cup) is equal to

$V=\pi r^{2}h$

we have

$r=36\ mm\\h=72\ mm$

substitute

$V=\pi (36)^{2}(72)$

$V=93,312\pi\ mm^3$ -----> exact value

<em>Find the approximate value</em>

assume

$\pi =3.14$

$V=93,312(3.14)=292,999.68\ mm^3$

7 0

3 years ago

PLEASE HELP!!! WILL MARK BRAINLIEST!! THX! A golfer hits a ball with an initial velocity of 32.7 m/s from the ground. Find the f

OLEGan [10]

Answer:

See below

Step-by-step explanation:

<u>First Problem</u>

The ball hits the ground when $h(t)=0$ , therefore:

$h(t)=-4.9t^2+v_0t+h_0$

$0=-4.9t^2+32.7t$

$0=t(-4.9t+32.7)$

$t=0$ and $t=\frac{32.7}{4.9}\approx6.67$

Since the ball is in the air before it hits the ground, $t=6.67$ (seconds) is the more appropriate choice.

<u>Second Problem</u>

The maximum height of the ball is determined when $t=-\frac{b}{2a}$ , therefore:

$t=-\frac{b}{2a}$

$t=-\frac{32.7}{2(-4.9)}$

$t=-\frac{32.7}{-9.8}$

$t\approx3.34$

This means that the height of the ball is at its maximum after 3.34 seconds:

$h(t)=-4.9t^2+32.7t$

$h(3.34)=-4.9(3.34)^2+32.7(3.34)$

$h(3.34)\approx54.55$

Thus, the answer is 54.55 (meters).

<u>Third Problem</u>

Refer to the second problem

<u>Fourth Problem</u>

<u /> $h(t)=-4.9t^2+32.7t$ <u />

<u /> $h(4.3)=-4.9(4.3)^2+32.7(4.3)$ <u />

<u /> $h(4.3)\approx50.01$ <u />

<u />

Therefore, the height of the ball after 4.3 seconds is 50.01 (meters).

<u>Fifth Problem</u>

The ball will be 24 meters off the ground when $h(t)=24$ , therefore:

$h(t)=-4.9t^2+32.7t$

$24=-4.9t^2+32.7t$

$0=-4.9t^2+32.7t-24$

$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$t=\frac{-32.7\pm\sqrt{(32.7)^2-4(-4.9)(-24)}}{2(-4.9)}$

$t_1\approx0.84$

$t_2\approx5.83$

Therefore, the ball will be 24 meters off the ground after 0.84 (seconds) and 5.83 (seconds)

5 0

2 years ago