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NikAS [45]
3 years ago
12

The sum of two numbers is 107, the difference is 51. Find the numbers. Answer : The numbers are

Mathematics
1 answer:
Fynjy0 [20]3 years ago
8 0

Answer:

The answer is 56.

Others are 50 + 57 = 107, 44 + 63 = 107, and 39,68

Step-by-step explanation:

107-51 = 56

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Look at the following pairs of numbers (x, y).
katrin2010 [14]

Step-by-step explanation:

Taking the first coordinate point (3,16.5)

where x= 3 and y= 16.5

\frac{y}{x}  =  \frac{16.5}{3}

\frac{y}{x}  = 5.5

y = 5.5x

optionB

8 0
3 years ago
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What is 5/8 written as a decimal and a percent?
grandymaker [24]
Im not sure about the percent but as a decimal its <em><u>1.6</u></em>
3 0
3 years ago
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Simplify the expression: 7d +12-14d-3
Akimi4 [234]
Group like terms
7d-14d+12-3
add them
-7d+9
6 0
3 years ago
A population of insects grows exponentially, as shown in the table. Suppose the increase in population continues at the same rat
Ivanshal [37]

We are told that a population of insects grows exponentially and we are given a table of data about insect population growth. We are asked to find population of insects at the end of week 11.        

The initial insect population is 20 and at the end of 1st week population increases to 30.

Let us find growth percentage of insect  population,

\text{Growth percentage}=\frac{\text{Difference}}{Actual} \cdot100

\text{Growth percentage}=\frac{\text{30-20}}{20} \cdot100

\text{Growth percentage}=\frac{\text{10}}{20} \cdot100

\text{Growth percentage}=0.5 \cdot100=50

We can see that insect population is growing at rate of 50% per week.

Now let us write an exponential function for our population.

P(w)=20(1+0.50)^{w}, where P(w) represents population at the end of w weeks.

Let us substitute w=11 in our function to find insect population at the end of 11 weeks.

P(11)=20(1+0.50)^{11}

P(11)=20(1.50)^{11}

P(11)=20\cdot 86.49755859375

P(11)=1729.951171875\approx 1730

Therefore, population of insects at the end of 11th week will be 1730.  



7 0
3 years ago
Please help me with this math problem.
liraira [26]

If we plug the values, we have

f(2) = 2p+3,\ g(5) = 5+4p

So, we have to solve the following equation:

2p+3 = 5+4p \iff -2p=2 \iff p = -1

So, when we impose f(x) = g(x), knowing the value of p, we have

f(x) = -x+3 = x-4 = g(x)

And so the solution is

-x+3 = x-4 \iff -2x = -7 \iff x = \dfrac{7}{2}

4 0
3 years ago
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