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2 years ago

2x - 3>5 what the answer

Mathematics

2 answers:

Zepler [3.9K]2 years ago

6 0

The answer is x>0.4
Just like what the other person said

Elden [556K]2 years ago

5 0

So first what you do is add 2x - 3 > 5
Why because do you remember PEMDAS well you have to do it so you have to do it the other way I mean SADMEP. You start with subtraction and you got 2
now what you got is 2x > 5 Now you do the opposite of multiple so you divide 2 divided by 5 is equal: x > 0.4

Hope I had helped

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2 years ago

Which expression is equivalent to<br> 5<br> 8

neonofarm [45]

Answer:

10/16

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4 0

1 year ago

Read 2 more answers

Lian is paid for SR 75 per hour. Using p for her pay and h for the hours of work, which function rule represents

Andru [333]

Answer:

The function p = 75h represents this situation.

Step-by-step explanation:

We know the equation of linear equation in the slope-intercept form is

$y = mx+b$

where

m = rate of change = slope

b = y-intercept

Let 'h' be the number of hours

Let 'p' be the pay

Given that Lian is paid SR 75 per hour.

so the rate of change of slope = m = 75

As there is no initial condition, so b = 0

y = mx+b

Now, mapping the information data into the slope-intercept form

p = y

m = 75

h = x

b = 0

Thus, the equation becomes

p = 75h

Therefore, the function p = 75h represents this situation.

VERIFICATION:

Given the equation

p = 75h

for h = 1

p = 75(1)

p = 75

Thus, Liam is paid SR 75 per hour.

4 0

2 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

2 years ago