find the volume of a cylinder with a radius of 12 cm and a height of 20 cm show work​

Answer:

x=3          x=-2

Step-by-step explanation:

(x-3)(x+2)=0

Using the zero product property

x-3 =0   x+2 =0

x=3          x=-2

Answer:

3 and -2

Step-by-step explanation:

x -3 =0

x= 3

x+2 =0

x= -2

Answer:  10

Step-by-step explanation:

Arranged data

7,10,10,11,11,12,12,13,13,17

Range= Highest value - Lowest Vlue

Highest value=17

Lowest value =7

Range=17-7

Range=10

Answer:

[tex]Answer:  10

Step-by-step explanation:

Arranged data

7,10,10,11,11,12,12,13,13,17

Range= Highest value - Lowest Vlue

Highest value=17

Lowest value =7

Range=17-7

Range=10 [/tex]

hope it is helpful to you

Answer:

42 + 42 + z = 180

Step-by-step explanation:

Note that:

Base angles in an isosceles triangle are equal in measure.

The sum of angles in a triangle is 180°.

Thus, the equation would be

42 + 42 + z = 180

whats the rest of the question?

Step-by-step explanation:

answer of number 5

length=7cm

breadth=9 cm

height=6 cm

volume=?

we know

v =l×b×h

=7×9×6

=378

so, the volume is 378 cm

Answer:

cone, radius is 3^2 x pi x 7/3 = 65.98

Step-by-step explanation:

Answer:

8,-1,-9,-18,-27,-36

and so on i would think it would be -27

Step-by-step explanation:

Answer:

-28

Step-by-step explanation:

other answer off by one

Answer:

44

Step-by-step explanation:

the steps are in the pic above

Step-by-step explanation:

a) unsuitable - the precision should have been about 0.030 kg (about an ounce)

b) unsuitable - the stopwatch should have a precision of 0.001 s

c) just about right, although I wouldn't mind it if its precision is of 0.05 mm

d) unsuitable - color matching requires much finer control so I would go with 1/8 cup precision.

Answer:

7

Step-by-step explanation:

210=5×6×x

x is length of aquarium.

210÷(5×6)=7

$/sqrt{9}$

Answer:

13

Step-by-step explanation:

Answer:

13

Step-by-step explanation:

48-[(2^2+3)x5]

48-[(4+3)x5]

48-[(7)x5]

48-[35]

13

Answer:

Solution given:

A triangle PQR is right angled at R, with hypotenuse{h}PQ=80cm

and

base[b]PR=60cm.

perpendicular [P]= QR

by using Pythagoras law

h²=p²+b²

80²=QR²+60²

QR²=80²-60²

QR=[tex]\sqrt{2800}[/tex]

QR=20[tex]\sqrt{7}[/tex]=52.9=53cm

QR=53cm.

Answer:

x = -3

Step-by-step explanation:

Given:

The population of animals = 350

Growth rate = 5% each year.

To find:

The exponential function to model this situation.

Solution:

The general exponential growth model is:

[tex]f(t)=a(1+r)^t[/tex]

Where, a is the initial value, r is the growth rate in decimal and t is the time period.

Substituting [tex]a=350, r=0.05[/tex] in the above formula, we get

[tex]f(t)=350(1+0.05)^t[/tex]

[tex]f(t)=350(1.05)^t[/tex]

Therefore, the required exponential function is [tex]f(t)=350(1.05)^t[/tex], where f(t) is the number of animals after t years.

Answer:

http://web.mnstate.edu/peil/MDEV102/U4/S36/S363.htmlStep-by-step explanation:

Answer:

First quartile: 35

Third Quartile: 46

Step-by-step explanation:

Answer:

48

Step-by-step explanation:

[tex]a = 12, \ b = 6, \ c = 3\\\\b^2 - c^2 + (a+ b+c) = 6^2 - 3^2 + (12 +6+3) = 36-9+21 = 48[/tex]

Answer:

2-12+(++)

b^{2}-1c^{2}+(a+b+c)

b2−1c2+(a+b+c)

Simplify: 1

Eliminate redundant parentheses

2−12+(++)

b^{2}-1c^{2}+(a+b+c)

b2−1c2+(a+b+c)

2−12+++

b^{2}-1c^{2}+a+b+c

b2−1c2+a+b+c

Solution: 2−12+++

b^{2}-1c^{2}+a+b+c

b2−1c2+a+b+c

Answer:

y>-2x-3

Step-by-step explanation:

I think, not sure.

Answer:

Step-by-step explanation:

3) (21²-6²)½ = 20.1 cm

4) (10²+7²)½ =12.2 cm

5) (6.5²-2.7²)½ = 5.9 cm

Answer:

third option mechanical weathering

Step-by-step explanation:

in mechanical weathering it breaks the rock into smaller pieces . these smaller pieces are just like the bigger rock , just smaller .

Given :

A motorboat travels 9 miles downstream (with the current) in 30 minutes.

The return trip upstream (against the current) takes 90 minutes.

To Find :

The speed of stream and motorboat.

Solution :

Let, speed of stream is u and speed of motorboat is v.

Now, we know distance is same in both case.

Distance( downstream ) = speed × time

[tex]9 = ( v +u ) \times 0.5\\\\v + u = 18[/tex]....1)

Distance( upstream ) = speed × time

[tex]9 = ( v- u ) \times 1.5\\\\v-u = 6[/tex]....2)

Solving equation 1 and 2 , we get :

v = 12 miles/hour and u = 6 miles/hour

Hence, this is the required solution.

Answer:

The speed of the stream is 6 mph, and the speed of the boat is 12 mph.

Step-by-step explanation:

Downstream:

9 miles in 30 minutes (0.5 hour).

9 mi / 0.5 h = 18 mph

Upstream:

9 miles in 90 minutes (1.5 h).

9 mi / 1.5 h = 6 mph

Let s = speed of stream.

The speed of the stream adds to the speed of the boat downstream and is subtracted fromt he speed of the boat upstream.

Let speed of boat be b.

b + s = 18

b - s = 6

Add the equations.

2b = 24

b = 12

b + s = 18

12 + s = 18

s = 6

The speed of the stream is 6 mph, and the speed of the boat is 12 mph.

Answer:

X is equal to 4

Answer:

a. 10.08

b. 0.0576

[tex]4.5[/tex] ✅

Step-by-step explanation:

[tex] \frac{1}{2} \times 9 \\ = \frac{1 \times 9}{2} \\ = \frac{9}{2} \\ = 4.5[/tex]

[tex]\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘[/tex]

Answer:

$509.97 is Rosa's regular pay

Answer:

4/ 5

cos A = AB/ AC

= 24/ 30

= 4/ 5

that is option A

Answer:

Area of length Vb = 25 square units

Area of segment Vb  = 26 square units

Step-by-step explanation:

Area of length Vb = theta/360 * 2πr  

Area of length Vb = 120/360 * 2*3.14 * 12  

Area of length Vb = 1/3 * 75.36  

Area of length Vb = 25 square units  

Ares of length Vb is 25.1 sq. units

Area of shaded portion = Area of sector VUB - Area of triangle VUB

Area of sector VUB = theta/360 * πr²

Area of sector VUB = 120/360 * 3.14*144

Area of sector VUB = 1/3 * 452.16

Area of sector VUB =150.72

Area of triangle VUB  = r²sintheta

Area of sector VUB = = 12²sin 120

Area of sector VUB = = 144(0.8660)

Area of sector VUB = 124.71

Area of the shaded segment = 150.72 - 124.71

Area of the shaded segment = 26.01

Area of the shaded segment = 26 sq. unitsAnswer:

Area of length Vb = 25 square units

Area of segment Vb  = 26 square units

Step-by-step explanation:

Area of length Vb = theta/360 * 2πr

Area of length Vb = 120/360 * 2*3.14 * 12

Area of length Vb = 1/3 * 75.36

Area of length Vb = 25 square units

Ares of length Vb is 25.1 sq. units

Area of shaded portion = Area of sector VUB - Area of triangle VUB

Area of sector VUB = theta/360 * πr²

Area of sector VUB = 120/360 * 3.14*144

Area of sector VUB = 1/3 * 452.16

Area of sector VUB =150.72

Area of triangle VUB  = r²sintheta

Area of sector VUB = = 12²sin 120

Area of sector VUB = = 144(0.8660)

Area of sector VUB = 124.71

Area of the shaded segment = 150.72 - 124.71

Area of the shaded segment = 26.01

Area of the shaded segment = 26 sq. units

Answer:

4 tens, 3 twenties

Step-by-step explanation:

4 tens  ($40) and 3 ($60)  twenties work. I’m sorry I don’t have a formula for it though.

Answer:

using pythagoras theorem

h^2=p^2+b^2

h^2=12^2+9^2

h^2=144+81

h^2=225

h=[tex]\sqrt{225}[/tex]

h=15

Step-by-step explanation:

Hi there!  

»»————-　★　————-««

I believe your answer is:  

[tex]\huge\boxed{2\sqrt{3}}[/tex]

»»————-　★　————-««  

Here’s why:  

[tex]\boxed{\text{Simplifying the square root of 12}}\\\\\rightarrow\sqrt{12}\\\\\text{4 can be factored out of 12.}\\\\\rightarrow\sqrt{4(3)}\\\\\text{Two '2's can be factored out of 4.}\\\\\rightarrow\sqrt{2*2*3}\\\\\rightarrow\sqrt{2^2(3)}\\\\\text{Pulling out the terms from under the radical...}\\\\\rightarrow\boxed{2\sqrt{3}}[/tex]

»»————-　★　————-««  

Hope this helps you. I apologize if it’s incorrect.  

Answer:

The initial height of the ball before launch is 393.12 feet.

Step-by-step explanation:

The ball's height ([tex]h[/tex]), in feet, as a function of time ([tex]t[/tex]), in seconds, is represented by [tex]h(t) = -16\cdot t^{2}+153.6\cdot t + 393.12[/tex]. The initial height of the ball before launch is the height of the ball evaluated at [tex]t = 0\,s[/tex]:

[tex]h(0) = -16\cdot (0)^{2}+153.6\cdot (0) + 393.12[/tex]

[tex]h(0) = 393.12\,ft[/tex]

The initial height of the ball before launch is 393.12 feet.

Answer:

C. 2/5 full

Step-by-step explanation:

[tex]\frac{1}{\frac{1}{15} } :\frac{6}{y}[/tex]

y · 1 = 1/15 · 6

y = 0.4

[tex]y=\frac{2}{5}[/tex]