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Nuetrik [128]
3 years ago
12

Find the missing length marked with a question mark. UVW-USR

Mathematics
1 answer:
mariarad [96]3 years ago
3 0

Given:

\Delta UVW\sim \Delta USR

To find:

The missing length marked with a question mark.

Solution:

Let x be the missing value.

We have,

\Delta UVW\sim \Delta USR

Corresponding sides of similar triangles are proportional, so

\dfrac{UV}{US}=\dfrac{VW}{SR}=\dfrac{UW}{UR}

Using this, we get

\dfrac{UV}{US}=\dfrac{UW}{UR}

On substituting the values, we get

\dfrac{70}{x}=\dfrac{55}{44}

\dfrac{70}{x}=\dfrac{5}{4}

On cross multiplication, we get

70\times 4=5\times x

280=5x

\dfrac{280}{5}=x

56=x

Therefore, the missing length is 56 units.

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5c-6=4-3c<br>2c-6=4<br>2c=10<br>c=5<br>describe the error made in the problem.​
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Answer:

5(c) - 6 = 4 - 3(c) is wrong

Step-by-step explanation:

The other problems are correct and equal. This one is wrong because when you do the equations you get 19 = -11 which is clearly not equal. You would need to change the equation to make it equal. So it would be:

5(c) - 6 = 4 - -3(c) so now it is 19 = 19.

So you would change 5(c) - 6 = 4 - 3(c)  into 5(c) - 6 = 4 - -3(c)

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3 years ago
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Solve for x: 3(x + 1)= -2(x - 1) + 6.
pickupchik [31]

Answer:

x=1

Step-by-step explanation:

3(x + 1)= -2(x - 1) + 6.

Distribute

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Add 2x to each side

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4 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%

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Answer:

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