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Diano4ka-milaya [45]
3 years ago
7

WILL GIVE BRIANLEIST DONT SCASM MEQuadrilaterals LMNO and STUV are similar. What is the value of x in inches? 6.6 in. 8.2 in. 22

.9 in. 27.1 in. In quadrilateral L M N O, the length of segment M N is two and two-tenths inches and the length of segment N O is four and one-tenth inches. In quadrilateral S T U V, the length of segment T U is unknown and represented by variable x, and the length of segment U V is twelve and three-tenths inches.

Mathematics
2 answers:
Klio2033 [76]3 years ago
7 0

Answer:

x = 6.6

Step-by-step explanation:

In order to solve the missing variable, we need to find the scale factor. I'll just take 12.3 and divide it by its corresponding side that is 4.1.

SF = 12.3/4.1 = 3

Now that we know the scale factor is 3, we can take 2.2 and multiply it by 3 to solve for x.

2.2(3) = 6.6

Therefore, x = 6.6.

jeka57 [31]3 years ago
6 0

Answer:

6.6 inches

Step-by-step explanation:

4.1 times 3 = 12.3

2.2 times 3 = 6.6

I think this is right... sorry if not

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A tank with a capacity of 1000 L is full of a mixture of water and chlorine with a concentration of 0.02 g of chlorine per liter
faltersainse [42]

At the start, the tank contains

(0.02 g/L) * (1000 L) = 20 g

of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.

Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s

In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.

So the amount of chlorine in the tank changes according to

\dfrac{\mathrm dc(t)}{\mathrm dt}=-\dfrac{5c(t)}{200-3t}

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):

\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0

\dfrac1{(200-3t)^{5/3}}\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{(200-3t)^{8/3}}=0

\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0

\dfrac{c(t)}{(200-3t)^{5/3}}=C

c(t)=C(200-3t)^{5/3}

There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :

20=C(200)^{5/3}\implies C=\dfrac1{200\cdot5^{1/3}}

\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}

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vovangra [49]

The answer is A

Step-by-step explanation:

I got is right pls mark me brainliest

3 0
3 years ago
Read 2 more answers
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