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anzhelika [568]
3 years ago
5

In triangle ABC,of angle A=angle B=45°,name the longest side.​

Mathematics
1 answer:
serg [7]3 years ago
3 0
The longest side would be C
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Use a calculator to evaluate sec 85 degrees
faltersainse [42]

OK.  I used my calculator to evaluate sec(85 degrees).

My calculator doesn't have a "sec" button on it. 
But I remembered that

           sec of an angle = 1 / (cosine of the same angle) .

So I used my calculator to find cos(85), and then I hit the
" 1/x " key, and got 11.474, which I knew to be sec(85).

3 0
3 years ago
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}}

7 0
3 years ago
Help!!!!!!!!!!!!!!!!!!!!!!!!!!1
Ugo [173]
Answer: D) y= 14/17x - 4/17
3 0
3 years ago
a cylindrical vase has a diameter of 6 inches at the bottom of the vase there are 9 marbles each of diameter 3 inches the vase i
KiRa [710]

Answer:

<h2>The volume of water in the vase is 339.33 in^3</h2>

Step-by-step explanation:

To calculate the volume of water in the vase we need the following parameters

1. the diameter/radius of the vase

2. the height /level of water in the vase

Given data

diameter d= 6 in

radius = d/2= 6/2 = 3 in

height of water h= 12 in

we know that the expression for the volume of a cylinder is given as

volume=  \pi r^2h

Inserting our data we have

volume= 3.142*3^2*12\\\volume= 3.142*9*12\\\volume= 339.33 in^3

3 0
3 years ago
3) g(x)= x3 + x<br> h(x) = x + 4<br> Find g(0)h(0)
Arte-miy333 [17]

Answer:

goh(x)=x^3+12x^2+49x+68

Step-by-step explanation:

g(x) = x^3+x

h(x)=x+4

We have to find

goh(x)

goh(x) = g(h(x))=g(x+4)

If g(x) = x^3+x

g(x+4) = (x+4)^3+(x+4)

           =x^3+12x^2+49x+68

Hence our answer is

goh(x)=x^3+12x^2+49x+68

4 0
3 years ago
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