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3 years ago

HELP PLEASE BECAUSE KHAN ACADEMY SUCKS!

Mathematics

2 answers:

Anastaziya [24]3 years ago

7 0

I think it’s 10 cm
Don’t know if I’m right tho

Andru [333]3 years ago

4 0

volume Large Can
r = 8 cm
pi = 3.14
h = 20 cm
V = pi r^2 h
V = 3.14*8^2 * 20
V = 4019 cm^3

Volume left over.
He has 1/4 of a can of paint left.
V1 = 1/4 * 4019
V1 = 1004.8

Calculate height.
V = 1004.8
pi = 3.14
r = 5 cm^2
h = ??

h = V/(pi r^2)
h = 1004.8 / (3.14 * 5^2)
h = 1004.8 / 78.5
h = 12.8 cm

There are other ways of doing this, but this is the clearest.

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The velocity of a ball is thrown up is 20 meters/second. It attains zero velocity after 3.5 seconds. What is the average acceler

Reptile [31]

Average acceleration: ( v - v o ) / Δ t
v = 0 m/s, v o = 20 m/s, Δ t = 3.5 s
a (average) = ( 0 m/s - 20 m/s )/3.5 s = - 20 m/s / 3.5 s =
= - 5.7142857 m/s² ≈ - 5.71 m/s²

7 0

4 years ago

Read 2 more answers

8a+6ab+5b,-6a-ab-8b and - 4a+2ab+3b<br><br>Pls answer this question

suter [353]

Answer:

4 a+8 ab+8b

Step-by-step explanation:

8 a-4 a+6ab+2ab+5 b+3 b

4 a+8 ab+8b

6 0

3 years ago

Help ...............

sammy [17]

A straight line is 180° so....

9x+24 +6x - 24 = 180

15x=180. ( because you have to combine like terms.

X= 12

5 0

4 years ago

SOLVE THE QUESTION BELOW ASAP

qwelly [4]

Answer:

Part A) The graph in the attached figure (see the explanation)

Part B) 16 feet

Part C) see the explanation

Step-by-step explanation:

Part A) Graph the function

Let

h(t) ----> the height in feet of the ball above the ground

t -----> the time in seconds

we have

$h(t)=-16t^{2}+98$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex is a maximum

To graph the parabola, find the vertex, the intercepts, and the axis of symmetry

<em>Find the vertex</em>

The function is written in vertex form

The vertex is the point (0,98)

Find the y-intercept

The y-intercept is the value of the function when the value of t is equal to zero

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98$

The y-intercept is the point (0,98)

Find the t-intercepts

The t-intercepts are the values of t when the value of the function is equal to zero

For h(t)=0

$-16t^{2}+98=0$

$t^{2}=\frac{98}{16}$

square root both sides

$t=\pm\frac{\sqrt{98}}{4}$

$t=\pm7\frac{\sqrt{2}}{4}$

therefore

The t-intercepts are

$(-7\frac{\sqrt{2}}{4},0), (7\frac{\sqrt{2}}{4},0)$

$(-2.475,0), (2.475,0)$

Find the axis of symmetry

The equation of the axis of symmetry in a vertical parabola is equal to the x-coordinate of the vertex

$x=0$ ----> the y-axis

To graph the parabola, plot the given points and connect them

we have

The vertex is the point $(0,98)$

The y-intercept is the point $(0,98)$

The t-intercepts are $(-2.475,0), (2.475,0)$

The axis of symmetry is the y-axis

The graph in the attached figure

Part B) How far is the artifact fallen from the time t=0 to time t=1

we know that

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98\ ft$

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

Find the difference

$98\ ft-82\ ft=16\ ft$

Part C) Does the artifact fall the same distance from time t=1 to time t=2 as it does from the time t=0 to time t=1?

we know that

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

For t=2

$h(t)=-16(2)^{2}+98$

$h(2)=34\ ft$

Find the difference

$82\ ft-34\ ft=48\ ft$

The artifact fall 48 feet from time t=1 to time t=2 and fall 16 feet from time t=0 to time t=1

therefore

The distance traveled from t=1 to t=2 is greater than the distance traveled from t=0 to t=1

8 0

3 years ago

(31 x 15) + (108 divided by 6)=

Solnce55 [7]

465+18 = 483
The fraction was just to throw you off, it's not too hard if you look carefully since you get a whole number from it

4 0

3 years ago

Read 2 more answers