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vichka [17]
3 years ago
9

Your class can fit no more than 20 students. Write an inequality that best fits this situation.

Mathematics
1 answer:
Wittaler [7]3 years ago
8 0

Answer:

number-of-students \leq 20

x\leq 20

You might be interested in
How do you solve 2(x+7)+x=20
alexira [117]
2(x+7) + x=20
(2)(x) + (2)(7) + x=20  Distribute 
2x+14 + x =20
(2x+x) + (14) =20  Combine Like Terms 
3x+14=20 
    - 14  -14             Subtract 14 from both sides 
3x = 6 
3x/3 6/3                  Divide Both Sides by 3 
 
x = 2 


Let me know if you still don't understand 

4 0
2 years ago
Given 4 c + 5 not-equals 0 and c is a real number, what is the multiplicative inverse of 4 c + 5 not-equals 0
Irina-Kira [14]

Answer:

1 / 4c+5

Step-by-step explanation:

Given :

4c + 5 ≠ 0

c is a real number.

Find the multiplicative inverse of equation ?

The multiplicative inverse of a number is reciprocal of number

Let

x = a number

The multiplicative inverse = 1/x

Let 4c + 5 ≠ 0 be a number x

Then, the multiplicative inverse of x = 1/x

x = 4c + 5

1/x = 1 / 4c+5

Therefore, the multiplicative inverse of 4c + 5 = 1 / 4c+5

7 0
3 years ago
Jill and Gary share £63 in the ratio 2:5<br> Work out how much each person gets.
irina [24]

Answer:

Jill gets €18 Gary gets €45.

Step-by-step explanation:

First you add the two numbers of the ratio together. 2+5=7. Then you divide the constant, 63, by the total of the ratio, 7. 63/7=9. Then you multiply each number of the ratio by the quotient. 2 x 9= 18. 5 x 9= 45.

4 0
3 years ago
Solve:
Temka [501]

Answer:

a= -5

Step-by-step explanation

PEMDAS

8 0
2 years ago
There are 7800 grams of lodine that have a half-life of 8 days. How much lodine will remain after 40 days?
anygoal [31]

Given that we know the initial mass of Iodine and its half-life, we want to see how much will remain after 40 days.

After 40 days, 244.17 grams of Iodine will remain.

<h3 /><h3>The half-life of materials and how to use it:</h3>

The half-life of a material is the time it takes for that amount of material to reduce to its half.

We can model the amount of Iodine as:

A(t) = A*e^{k*t}

  • Where A is the initial amount, in this case, 7800g.
  • k is a constant that depends on the half-life.
  • t is the time in days.

Replacing what we know, we get:

A(t) = 7800g*e^{k*t}

Now we use the fact that the half-life is 8 days, this means that:

e^{k*8} = 1/2

ln(e^{k*8}) = ln(1/2)

k*8 = ln(1/2)

k = ln(1/2)/8 = -0.0866

Then the function is:

A(t) =  7800g*e^{-0.0866*t}

So now we just need to evaluate this in t = 40.

A(40) = 7800g*e^{-0.0866*40} = 244.17g

So, after 40 days, 244.17 grams of Iodine will remain.

If you want to learn more about half-life and decays, you can read:

brainly.com/question/11152793

3 0
2 years ago
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