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egoroff_w [7]
3 years ago
10

John earned three dollars for each lawn he mowed. If he had eight lawns to mow, but forget mow two of them, how much money did h

e actually earn?
Mathematics
1 answer:
gayaneshka [121]3 years ago
4 0
If John had to mow 8 lawns, but he forgot to mow 2 lawns, we have to subtract 2 from 8 to see how many lawns he actually mowed.

8-2=6

John mowed 6 lawns!

For each lawn John mowed, he earned 3 dollars. So, we have to multiply the number of lawns he mowed, 6, by the amount of dollars he earned per lawn, 3. Multiplying 6*3 is the same as adding 3+3+3+3+3+3!

6*3=18

John earned $18!

Hope I helped! :)

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Simplify 3 x (7 - 2) + 6
erastovalidia [21]

Answer: 1

Step-by-step explanation:

8/-4 ÷ -3/9

8/-4 × 9/ -3 = 6

5 0
2 years ago
Determine whether the points (–3,–2) and (3,2) are in the solution set of the system of inequalities below. y ≤ ½x + 2 y < –2
lana [24]

Given:

The system of inequalities:

y\leq \dfrac{1}{2}x+2

y

To find:

Whether the points (–3,–2) and (3,2) are in the solution set of the given system of inequalities.

Solution:

A point is in the solution set of the given system of inequalities if it satisfies both inequalities.

Check for the point (-3,-2).

-2\leq \dfrac{1}{2}(-3)+2

-2\leq -1.5+2

-2\leq 0.5

This statement is true.

-2

-2

-2

This statement is also true.

Since the point (-3,-2) satisfies both inequalities, therefore (-3,-2) is in the solution set of the given system of inequalities.

Now, check for the point (3,2).

2

2

2

This statement is false because 2>-9.

Since the point (3,2) does not satisfy the second inequality, therefore (3,2) is not in the solution set of the given system of inequalities.

7 0
3 years ago
Show both decimal places (5.06).
lilavasa [31]

a. the cost of a call for six minutes is 0.70.

b.the cost of a 14-minute call is 2.62.

c.the cost of a 9 ½2-minute call is 1.66.

<h3>How to find the cost of each call ?</h3>

given that

rate schedule for an m-minute call from any of its pay phones 0.70.

c(m) = 0.70 when m<=6

c(m) = 0.70+ 0.24(m-6) when m>6 & m is an integer

c(m) = 0.70+0.24((m-6)+1) when m>6 & m is not an integer.

a. to find the cost of a call for six minutes.

already given that c(m) =0.70 when m=6.

Therefore, the cost of a call for six minutes is 0.70.

b. to find the cost of a 14-minute call.

according to given information

c(14) = 0.70 + 0.24(m - 6) \\ c(14) = 0.70 + 0.24(14 \: - 6)  \\c(14) = 0.70 + (0.24 \times 14) - (0.24 \times 6) \\c(14) = 0.70 + (3.36 - 1.44) \\ c(14) = 0.70 + 1.92 \\ c(14) = 2.62

Therefore,the cost of a 14-minute call is 2.62.

c.to find the cost of a 9 ½2-minute call

according to given information

c(9  \frac{1}{2} 2) = 0.70 + 0.24((9 \frac{1}{2} 2 - 6) + 1) \\ c(9  \frac{1}{2} 2) = 0.70 + 0.24((9 - 6) + 1) \\ c(9  \frac{1}{2} 2) = 0.70 + 0.24(3+ 1) \\ c(9  \frac{1}{2} 2) = 0.70 + 0.24(4) \\ c(9  \frac{1}{2} 2) = 0.70 + 0.96 \\ c(9  \frac{1}{2} 2) = 1.66

Therefore,the cost of a 9 ½2-minute call is 1.66.

Learn more about problems on cost of call, refer:

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4 0
1 year ago
During normal breathing, about 12% of the air in the
nekit [7.7K]

The original air present after 8 breaths will be equal to 179.81 ml.

<h3>What is an equation?</h3>

The equation in mathematics is the relationship between the variables and the number and establishes the relationship between the two or more variables.

It is given in the question that:-

During normal breathing, about 12% of the air in the lungs is replaced after one breath

If the initial amount of air in the lungs is 500 mL

how much of the original air is present after 8 breaths?

The calculation for the eduction of the air will be done by the following equation:-

\rm =Volume \ present\times (0.88)^n\\\\\\= 500\times (0.88)^8

=179.88 ml

Hence original air present after 8 breaths will be equal to 179.81 ml.

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7 0
2 years ago
At a military function, three cannons are fired at intervals of 12
Morgarella [4.7K]

Answer:

144 seconds

Step-by-step explanation:

To do solve this question, we must find the LCM (least common multiple) of 12, 16, and 18.

Let's start by prime factorizing (finding the prime factors) all the numbers.

12\rightarrow3*4\rightarrow2*2*3

16\rightarrow4*4\rightarrow2*2*2*2

18\rightarrow2*9\rightarrow2*3*3

Now, we must find the greatest quantity of each [prime] number and multiply them together to obtain the least common multiple.

Upon further evaluation of the [prime] factors, we can see that we need four 2s and two 3s. Therefore...

2*2*2*2*3*3=2^4*3^2=16*9=144

They will be fired together again after 144 seconds.

3 0
2 years ago
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