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Damm [24]
3 years ago
7

Prove that angle ABD is congruent to angle CBE with solution!

Mathematics
1 answer:
WINSTONCH [101]3 years ago
8 0

ANSWER:

the conditions are the angle a is equal to angle c and ab = bc . Hence we need to prove that the triangles is congruent to the triangle cbe. ... angle A =angle C and AB=BC.

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Help asap please for brainlist !!!!!!!!!
IgorLugansk [536]
Bff i answered this earlier. It’s step 2!!! Because if you’re flipping the numerator and denominator, the sign of the exponent should be flipped as well. So the exponent will turn into -5x
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3 0
2 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
The length of a rectangle is 5 m less than twice the width, and the area of the rectangle is 52 m^2. Find the dimensions of the
lys-0071 [83]

length is 8m

Width is 6.5m

<u>Explanation:</u>

Let, width = b

So, length = 2b-5

area = 52m^{2}

We know,

area of rectangle = length X width

So, 52m^{2} = (2b-5) X b

On solving the above equation, we get

Width, b = 6.5m

length, l = 8m

4 0
3 years ago
Which shape has 2 parallel sides /i ready question
sesenic [268]

Answer:

A Parallelogram

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Holly cuts 2 ribbons into tenths for a craft project how many 1/10 size ribbons does she have
boyakko [2]

Answer:

Holly would have 20 1/10 sized ribbons.

Step-by-step explanation:

One whole ribbon would equal out to 10, this number multiplied by 2 would be 20. Therefore meaning she should have 20 1/10 sized pieces of ribbon.

6 0
3 years ago
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