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3 years ago

13

what is the maximum volume of water a hamster bath could hold with a depth of 1 2/3 inches, a length of 2 1/3 inches, and width

of 2 inches?

2 answers:

adoni [48]3 years ago

8 0

1 2/3 = 5/3

2 1/3 = 7/3

2 = 2

5/3 * 7/3 * 2 = 70/9 cubic inches = 0.127455 liters

Gennadij [26K]3 years ago

8 0

7 7/9 inches squared

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12. (05.02 LC) Look at the figure below: an image of a right triangle is shown with an angle labeled y If sin y° = 7 over q and

meriva

Answer:

$sec~y=\frac{q}{r}$

Step-by-step explanation:

$tan ~y=\frac{7}{r} \\\\\frac{sin~y}{cos~y} =\frac{7}{r} \\\\sin~y=\frac{7}{r} cos~y\\sin ~y=\frac{7}{q} \\\frac{7}{q} =\frac{7}{r} cos~y\\sec~y=\frac{7}{r} \times \frac{q}{7} =\frac{q}{r}$

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3 years ago

Find the value of x.

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Answer:

x = 36°

Step-by-step explanation:

Where a transversal crosses parallel lines, any obtuse angle it creates is supplementary to any acute angle it creates:

3x +2x = 180°

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x = 180°/5 = 36° . . . . . divide by 5

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2 years ago

Isosceles Right Triangle Reflection to prove ASA Congruence

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3 years ago

Consider the two triangles. Triangles A B C and H G I are shown. Angles A C B and H I G are right angles. The length of side A C

zheka24 [161]

Answer:

Both are similar by SAS similarity.

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Step-by-step explanation:

Step 1:

To prove that ACB and HIG as similar triangles.

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We have

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2 years ago

Read 2 more answers

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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3 years ago