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qwelly [4]
2 years ago
9

Five double-digit numbers We have five consecutive positive double-digit integers. If we swap places on the numbers in the large

st number, the sum of the five numbers increases by 1 more than the mean of the original numbers. What is the lowest number?
Mathematics
1 answer:
barxatty [35]2 years ago
3 0

Answer:

  • 33

Step-by-step explanation:

<u>Let the numbers are:</u>

  • x, x + 1, x + 2, x + 3 and x + 4
  • Their sum is 5x + 10
  • Their mean is (5x + 10)/2 = x + 2

If we swap the digits on the largest number, the sum increases by 1 more than x + 2

<u>We are looking for the number x + 4 = ab such that:</u>

  • ba - ab = ab - 2 + 1
  • ba = 2ab - 1
  • 10b + a = 2(10a + b) - 1
  • 10b + a = 20a + 2b - 1
  • 8b = 19a - 1
  • 19a = 8b + 1

<u>By trial method we get the solution:</u>

  • a = 3, b = 7

<u>Since x + 4 = 37, the lowest number is:</u>

  • x = 37 - 4 = 33

<u>Lets verify:</u>

  • 33, 34, 35, 36, 37

<u>The sum is:</u>

  • 5*33 + 10 = 175

<u>The mean is:</u>

  • 35

<u>Change the largest number tp 73 and find the sum again:</u>

  • 175 + (73 - 37) = 175 + 36 = 175 + 35 + 1

The sum has increased by 1 more than 35

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Michelle pays $66.85 for a costume pattern and 8 yards of fabric. The costume pattern cost $4.85 how much does each yard of fabr
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Step-by-step explanation:

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3 years ago
A hackberry tree has roots that reach a depth of
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Answer:

24.6967 meters

Step-by-step explanation:

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3 years ago
Evaluate (x + y)0 for x = -3 and y = 5.<br><br> a.) -1/2<br> b.) 0<br> c.) 1<br> d.) 2
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8 0
4 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%2Bx-3%29%3A%20%28x%5E%7B2%7D%20-4%29%5Cgeq%201" id="TexFormula1" title="(x^{
Jlenok [28]

Answer:

x>2

Step-by-step explanation:

When given the following inequality;

(x^2+x-3):(x^2-4)\geq1

Rewrite in a fractional form so that it is easier to work with. Remember, a ratio is another way of expressing a fraction where the first term is the numerator (value over the fraction) and the second is the denominator(value under the fraction);

\frac{x^2+x-3}{x^2-4}\geq1

Now bring all of the terms to one side so that the other side is just a zero, use the idea of inverse operations to achieve this:

\frac{x^2+x-3}{x^2-4}-1\geq0

Convert the (1) to have the like denominator as the other term on the left side. Keep in mind, any term over itself is equal to (1);

\frac{x^2+x-3}{x^2-4}-\frac{x^2-4}{x^2-4}\geq0

Perform the operation on the other side distribute the negative sign and combine like terms;

\frac{(x^2+x-3)-(x^2-4)}{x^2-4}\geq0\\\\\frac{x^2+x-3-x^2+4}{x^2-4}\geq0\\\\\frac{x+1}{x^2-4}\geq0

Factor the equation so that one can find the intervales where the inequality is true;

\frac{x+1}{(x-2)(x+2)}\geq0

Solve to find the intervales when the equation is true. These intervales are the spaces between the zeros. The zeros of the inequality can be found using the zero product property (which states that any number times zero equals zero), these zeros are as follows;

-1, 2, -2

Therefore the intervales are the following, remember, the denominator cannot be zero, therefore some zeros are not included in the domain

x\leq-2\\-2

Substitute a value in these intervales to find out if the inequality is positive or negative, if it is positive then the interval is a solution, if it is negative then it is not a solution. This is because the inequality is greater than or equal to zero;

x\leq-2   -> negative

-2   -> neagtive

-1\leq x   -> neagtive

x>2   -> positive

Therefore, the solution to the inequality is the following;

x>2

6 0
3 years ago
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