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3 years ago

Pleaassseeeee helpppppp,,,,!!!!!!

Mathematics

1 answer:

Flura [38]3 years ago

6 0

Surface Area (SA) of a Cylinder formula:
SA = 2*pi*r*h+2*pi*r^2
If you fill in the formula:
SA = 2*3.14*2*9+2*3.14*2^2
Simplify:
SA = 6.28*18+6.28*4
SA = 113.04 + 25.12
SA = 138.16 in
I hope I helped
If its wrong I'm super duper extra sorry!

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Semmy [17]

Step-by-step explanation:

y=75+50(36 hours)=1875

give me brainliest

4 0

2 years ago

Three Texas drovers met on a highway and began to dicker as follows:

Alik [6]

Answer:

Jim's: 7
Duke's: 21
Hank's: 11

Step-by-step explanation:

Let J, D, and H represent the drove sizes of Jim, Duke, and Hank.

2(H -6 + 1) = (J +6 -1) . . . . . Hank's offer

3(D -14 +1) = (H +14 -1) . . . . Duke's offer

6(J -4 +1) = (D +4 -1) . . . . . . Jim's offer

These equations can be simplified to ...

2H - J = 15

3D - H = 52

6J - D = 21

<u>Solution</u>

Adding twice the second equation to the first gives ...

2(3D -H) +(2H -J) = 2(52) +(15)

6D -J = 119

Adding 6 times the third equation to this one gives ...

6(6J -D) +(6D -J) = 6(21) +(119)

35J = 245

J = 7

Using this value in the third equation gives ...

6(7) -D = 21

42 -21 = D = 21

Using the value of J in the first equation gives ...

2H -7 = 15

H = (15 +7)/2 = 11

Then the solution is (J, D, H) = (7, 21, 11).

Jim's drove had 7 animals; Duke's drove had 21 animals; Hank's drove had 11 animals.

5 0

3 years ago

True or False

kolezko [41]

True Hope this helps! ;D

5 0

3 years ago

Item 3 The Hampton family used 17,158 kilowatts of electricity last year. They used about the same amount of electricity each we

katen-ka-za [31]

17,158/52= 329.96 rounded to the nearest tenth is 330

7 0

2 years ago

Find the area each sector. Do Not round. Part 1. NO LINKS!!<br><br>

sladkih [1.3K]

Answer:

$\textsf{Area of a sector (angle in degrees)}=\dfrac{\theta}{360 \textdegree}\pi r^2$

$\textsf{Area of a sector (angle in radians)}=\dfrac12r^2\theta$

17) Given:

$\theta$ = 240°
r = 16 ft

$\textsf{Area of a sector}=\dfrac{240}{360}\pi \cdot 16^2=\dfrac{512}{3}\pi \textsf{ ft}^2$

19) Given:

$\theta=\dfrac{3 \pi}{2}$
r = 14 cm

$\textsf{Area of a sector}=\dfrac12\cdot14^2 \cdot \dfrac{3\pi}{2}=147 \pi \textsf{ cm}^2$

21) Given:

$\theta=\dfrac{ \pi}{2}$
r = 10 mi

$\textsf{Area of a sector}=\dfrac12\cdot10^2 \cdot \dfrac{\pi}{2}=25 \pi \textsf{ mi}^2$

23) Given:

$\theta$ = 60°
r = 7 km

$\textsf{Area of a sector}=\dfrac{60}{360}\pi \cdot 7^2=\dfrac{49}{6}\pi \textsf{ km}^2$

3 0

2 years ago