All categories

2 years ago

PLEASE HELP! ON A TEST! (Show work)

Mathematics

1 answer:

Fiesta28 [93]2 years ago

6 0

Answer:

Um what is the question on the test?

You might be interested in

Find the volume of a right circular cone that has a height of 11.4 in and a base with a radius of 5.1 in. Round your answer to t

ella [17]

Answer:

Step-by-step explanation:

h=11.4 in

r=5.1 in

v=3.14*r^2*(h/3)

V=3.14*5.1^2*(11.4/3)=3.14*26.01*3.8=310.35

5 0

2 years ago

William works at a nearby electronics store. He makes a commission of 6% percent on everything he sells. If he sells a computer

Naddika [18.5K]

Commission earned by William is $ 45.84

<h3><u>Solution:</u></h3>

Given that,

William makes a commission of 6% percent on everything he sells

He sells a computer for $764.00

<em><u>To find: Commission amount of William</u></em>

Given that he makes a commission of 6 % on everything he sells. So he has received a commission of 6 % of $ 764.00

Commission amount of William = 6 % of $ 764.00

a % of b can be written in fraction as $\frac{a}{100} \times b$

$6 \% \text { of } 764=\frac{6}{100} \times 764=0.06 \times 764=45.84$

Thus commission earned by William is $ 45.84

8 0

3 years ago

A student on a piano stool rotates freely with an angular speed of 2.85 rev/s . The student holds a 1.50 kg mass in each outstre

Vlad1618 [11]

Answer:r'=0.327 m

Step-by-step explanation:

Given

$N=2.85 rev/s$

angular velocity $\omega =2\pi N=17.90 rad/s$

mass of objects $m=1.5 kg$

distance of objects from stool $r_1=0.789 m$

Combined moment of inertia of stool and student $=5.53 kg.m^2$

Now student pull off his hands so as to increase its speed to 3.60 rev/s

$\omega _2=2\pi N_2$

$\omega _2=2\pi 3.6=22.62 rad/s$

Initial moment of inertia of two masses $I_0=2mr_^2$

$I_0=2\times 1.5\times (0.789)^2=1.867$

After Pulling off hands so that r' is the distance of masses from stool

$I_0'=2\times 1.5\times (r')^2$

Conserving angular momentum

$I_1\omega =I_2\omega _2$

$(5.53+1.867)\cdot 17.90=(5.53+I_o')\cdot 22.62$

$I_0'=1.397\times 0.791$

$I_0'=5.851$

$5.53+2\times 1.5\times (r')^2=5.851$

$2\times 1.5\times (r')^2=0.321$

$r'^2=0.107009$

$r'=0.327 m$

7 0

3 years ago

Please help expand the expression using pascal's triangle<br><br> (1-2x)^4

aliya0001 [1]

(1 - 2x)⁴
(1 - 2x)(1 - 2x)(1 - 2x)(1 - 2x)
[1(1 - 2x) - 2x(1 - 2x)][1(1 - 2x) - 2x(1 - 2x)]
[1(1) - 1(2x) - 2x(1) - 2x(-2x)][1(1) - 1(2x) - 2x(1) - 2x(-2x)]
(1 - 2x - 2x + 4x²)(1 - 2x - 2x + 4x²)
(1 - 4x + 4x²)(1 - 4x + 4x²)
1(1 - 4x + 4x²) - 4x(1 - 4x + 4x²) + 4x²(1 - 4x + 4x²)
1(1) - 1(4x) + 1(4x²) - 4x(1) - 4x(-4x) - 4x(4x²) + 4x²(1) - 4x²(4x) + 4x²(4x²)
1 - 4x + 4x² - 4x + 16x² - 16x³ + 4x² - 16x³ + 16x⁴
1 - 4x - 4x + 4x² + 16x² + 4x² - 16x³ - 16x³ + 16x⁴
1 - 8x + 24x² - 32x³ + 16x⁴

5 0

2 years ago

Read 2 more answers

Solve for r. -49 = -7r

Nikolay [14]

R=7 because you would do -49 divided by -7

5 0

1 year ago