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3 years ago

1: What is the length of PL?

Mathematics

1 answer:

Yuliya22 [10]3 years ago

6 0

Answer:

PL 4

GL 12

HJ 10

Step-by-step explanation:

PL pythagoras theorem

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Find the value of x<br><br> 60°<br> 80°<br> 70°

klemol [59]

Answer:

that is 70° I hope you get a A+

6 0

3 years ago

A tank contains 60 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

MissTica

Answer:

(a) 60 kg; (b) 21.6 kg; (c) 0 kg/L

Step-by-step explanation:

(a) Initial amount of salt in tank

The tank initially contains 60 kg of salt.

(b) Amount of salt after 4.5 h

$\text{Let A = mass of salt after t min}\\\text{and }r_{i} = \text{rate of salt coming into tank}\\\text{and }r_{0} =\text{rate of salt going out of tank}$

(i) Set up an expression for the rate of change of salt concentration.

$\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\\text{The fresh water is entering with no salt, so}\\ r_{i} = 0\\r_{o} = \dfrac{\text{3 L}}{\text{1 min}} \times \dfrac {A\text{ kg}}{\text{1000 L}} =\dfrac{3A}{1000}\text{ kg/min}\\\\\dfrac{\text{d}A}{\text{d}t} = -0.003A \text{ kg/min}$

(ii) Integrate the expression

$\dfrac{\text{d}A}{\text{d}t} = -0.003A\\\\\dfrac{\text{d}A}{A} = -0.003\text{d}t\\\\\int \dfrac{\text{d}A}{A} = -\int 0.003\text{d}t\\\\\ln A = -0.003t + C$

(iii) Find the constant of integration

$\ln A = -0.003t + C\\\text{At t = 0, A = 60 kg/1000 L = 0.060 kg/L} \\\ln (0.060) = -0.003\times0 + C\\C = \ln(0.060)$

(iv) Solve for A as a function of time.

$\text{The integrated rate expression is}\\\ln A = -0.003t + \ln(0.060)\\\text{Solve for } A\\A = 0.060e^{-0.003t}$

(v) Calculate the amount of salt after 4.5 h

a. Convert hours to minutes

$\text{Time} = \text{4.5 h} \times \dfrac{\text{60 min}}{\text{1h}} = \text{270 min}$

b.Calculate the concentration

$A = 0.060e^{-0.003t} = 0.060e^{-0.003\times270} = 0.060e^{-0.81} = 0.060 \times 0.445 = \text{0.0267 kg/L}$

c. Calculate the volume

The tank has been filling at 6 L/min and draining at 3 L/min, so it is filling at a net rate of 3 L/min.

The volume added in 4.5 h is

$\text{Volume added} = \text{270 min} \times \dfrac{\text{3 L}}{\text{1 min}} = \text{810 L}$

Total volume in tank = 1000 L + 810 L = 1810 L

d. Calculate the mass of salt in the tank

$\text{Mass of salt in tank } = \text{1810 L} \times \dfrac{\text{0.0267 kg}}{\text{1 L}} = \textbf{21.6 kg}$

(c) Concentration at infinite time

$\text{As t $\longrightarrow \, -\infty,\, e^{-\infty} \longrightarrow \, 0$, so A $\longrightarrow \, 0$.}$

This makes sense, because the salt is continuously being flushed out by the fresh water coming in.

The graph below shows how the concentration of salt varies with time.

3 0

3 years ago

Which of the following has no solution?

tresset_1 [31]

Answer:

$(x+11)$

Step-by-step explanation:

Solve all options.

1. $(x+1$

$x+1$

So,

$(x+1$

There are solutions.

2. $(x+1\le1)\cap (x+1\ge 1)$

$x+1\le 1\Rightarrow x\le 0\\ \\x+1\ge 1\Rightarrow x\ge 0$

So,

$(x+1\le 1)\cap (x+1\ge 1)\Rightarrow (x\le 0)\cap (x\ge 0)\Rightarrow (x=0)$

There is one solution.

3. $(x+11)$

$x+10$

So,

$(x+11)\Rightarrow (x0)\Rightarrow (x\in \emptyset)$

There are no solutions.

7 0

3 years ago

Read 2 more answers

Factorise x squared - 3x

stepladder [879]

Answer:

x squared -3× =x(x-3)

Step-by-step explanation:

Since the common factor is x,one comes out to multiply the other to give it x squared and since one x has been written outside already we just write the 3 so that when we multiply it we will get the answer we were asked to factorise

7 0

3 years ago

$875 at 9.5% compunded twice a year. How much money will there be in 5years?

TEA [102]

P = A(1+i)ⁿ, where:
A= initial value
I = interest in %
n= period of compound investment.
However, if the interest is paid twice a year, that means you have to divide the interest by 2 and the period becomes twice n
P = 875(1+9.5%/2)⁵ˣ²
P = 875(1+4.75%)¹⁰
P =875( 1.0475)¹⁰
P = $1,391.7

5 0

4 years ago