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3 years ago

8

Fifteen sprinters ran the 50-yard dash in an average of 9.5 seconds. Of those runners,3 ran the dash in 11.5 seconds. Find the a

verage time of the other 12 boys

2 answers:

Blizzard [7]3 years ago

8 0

Various answer 4 twelve boys

dezoksy [38]3 years ago

3 0

There might be a variety of averages for the 12 boys.

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Distribute and the combine like terms 5-3(n+4) -2

murzikaleks [220]

The answer is 3(-3-n) :) hope this helps

3 0

3 years ago

4x-5(x-2)=12x+4-x<br> ??????????????

Anestetic [448]

   4x - 5(x - 2) = 12x + 4 - x
4x - 5(x) - 5(-2) = 12 - x + 4
    4x - 5x + 10 = 11x + 4
      -x + 10 = 11x + 4
      <u>+ x    + x    </u>
      10 = 12x + 4
      <u>- 4 - 4</u>
      <u>6</u> = <u>12x</u>
   12 12
   0.5 = x

6 0

3 years ago

Find the volume and surface area of the composite figure. Give your answer in terms of π.

irakobra [83]

Answer:

V = 240π cm^3 , S= 168π cm^2

Step-by-step explanation:

The given figure is a combination of hemi-sphere and a cone

<u>Volume:</u>

For volume

r = 6 cm

h = 8 cm

$Volume\ of\ cone = \frac{1}{3}\pi r^2h\\= \frac{1}{3}\pi (6)^2*8\\=\frac{1}{3}\pi *36*8\\=\frac{288}{3}\pi\\=96\pi cm^3 \\\\Volume\ of\ hemisphere = \frac{2}{3}\pi r^3\\=\frac{2}{3}*\pi * (6)^3\\=\frac{2}{3}*\pi *216\\=\frac{432}{3}\\=144\pi cm^3 \\\\Total\ Volume= Volume\ of\ cone + Volume\ of\ hemisphere\\= 96\pi +144\pi \\=240\pi cm^3$

<u>Surface Area:</u>

For this particular figure we have to consider the lateral area of the cone shape and surface area of the hemisphere

We have to find the lateral height

$l = \sqrt{r^2+h^2}\\ l = \sqrt{(6)^2+(8)^2} \\l= \sqrt{36+64}\\ l = \sqrt{100}\\l = 10cm\\\\Surface\ area\ of\ cone = \pi rl\\= \pi (6)(10)\\=\pi *60\\=60 \pi\ cm^2\\\\Surface\ area\ of\ hemisphere = 2\pi r^2\\= 2 \pi * (6)\\= 2 \pi *36\\= 72 \pi\ cm^2\\\\Total\ surface\ Area = Surface\ area\ of\ cone + Surface\ area\ of \ hemisphere\\= 60 \pi + 72 \pi\\=132 \pi\ cm^2$

Hence the first option is correct ..

3 0

3 years ago

What is the equation of the line in slope-intercept form?

inessss [21]

Answer:

y = x - 9

Step-by-step explanation:

the line has a slope of 1 so you do not have to include it in the equation and it crosses the y-axis at -9 so the y-intercept is -9

7 0

3 years ago

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m

borishaifa [10]

Answer:

the concentration at 10 minutes= 0.4+0.0133= 0.4133%

Step-by-step explanation:

Air containing 0.04% carbon dioxide

V, volume of room is 6000 ft3.

Q, rate of air 2000 ft3/min,

initial concentration of 0.4% carbon dioxide,

determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes?

firstly, we find the time taken for air to completely filled the room

Q = V/t

t = V/Q = 6000/2000 = 3min

so, its take 3mins for air to be completely filled in the room and for exhaust air to move out.

there is an initial concentration of 0.4% carbon dioxide, and the air pump in is 0.04%.

therefore,

3mins = 0.04% of CO2

3*60 =180sec = 0.04%

1sec = 0.04/180 = 0.00022%/sec

so at any time the concentration of CO2 is 0.4 + 0.00022 =0.40022%/sec

What is the concentration at 10 minute

the concentration at 10minutes = the concentration for 1minute because at every minutes, the concentration moves in is moves out. = concentration for 2000ft3.

for 0.04% = 6000ft3

? = 2000ft3

= 2000* 0.04)/6000 =0.0133%

the concentration at 10 minutes= 0.4+0.0133= 0.4133%

4 0

3 years ago