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timama [110]
9 months ago
12

BRAINLIEST !! Evaluate the following fractions giving your awnser in its simplest form.

Mathematics
2 answers:
kiruha [24]9 months ago
8 0
Answers are:
a) 1/15
b) 18
c) 1/12

a) 2/5 divided by 6
The rule in dividing fractions is to flip the fraction and multiply. Since “6” is 6/1, it becomes 1/6.
2/5 x 1/6 = 1/15

b) 3 1/3 times 5 2/5
Change the fractions into mixed fraction and multiply across.
10/3 x 27/5 = 270/15
Simplify by dividing numerator by denominator
270/15 = 18

c) 1/4 + 1/3 - 1/2
To add and subtract fractions the denominator has to be the same.
The lowest common denominator will be 12.
1/4 = 3/12
1/3 = 4/12
1/2 = 6/12

Now solve
3/12 + 4/12 - 6/12 = 1/12
Pie9 months ago
7 0

Answer:

question no c 0•08

Step-by-step explanation:

at first

1/4+1/3-1/2

0•25+0.33-0.50

0.58-0.50

0.08 answer

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Quadrilateral ABCD is a a square and the length of BE is 6 cm.<br><br>What is the length of AC?
Aloiza [94]

The answer is: 12 centimeters.

The explanation is shown below:

1. By definition, all four sides of the square are equal and the diagonals are equal too.

2. Keeping the information above on mind, you know that AC and BD are equal:

AC=BD

3. Then, the length of BD is twice BE. So, you can calculate it as following:

BD=2BE

BD=2(6cm)\\BD=12cm

4. Therefore, the lenght of AC is:

AC=12cm

5 0
3 years ago
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a
Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q  = -t/100 + c

Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c}  = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}

when t = 0, Q = 200 L × 1 g/L = 200 g

Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

2 = 200e^{(-t/100)}\\\frac{2}{200} =  e^{(-t/100)}

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0
3 years ago
4x - 5y = 15<br> find the slope and y-intercept
Mazyrski [523]

Answer:

slope=4/5

y-intercept = -3

Step-by-step explanation:

first we need to rewrite the expression to

y=mx+b

were b is the y-intercept and m the slope

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so there we have it

m=\frac{4}{5} \\\\b=-3

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Step-by-step explanation:

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