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Angelina_Jolie [31]
3 years ago
15

If the sum of five consecutive even integers is 90, what is the smallest of the five integers?

Mathematics
1 answer:
ruslelena [56]3 years ago
5 0

So the pattern of even integers is: 2, 4, 6, 8, etc. They are always 2 more than the prior even integer. Using this knowledge, our 5 consecutive even numbers are (let x = smallest even number): x, x + 2, x + 4, x + 6, and x + 8.

Now that we have the numbers, we can form our equation as such: x+x+2+x+4+x+6+x+8=90 . From here we can solve for x.

Firstly, combine like terms: 5x+20=90

Next, subtract 20 onto both sides of the equation: 5x=70

Lastly, divide both sides by 5 and <u>your smallest integer is x=14</u>

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In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini
svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2

The augmented matrix is

\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]

The reduction of this matrix to row-echelon form is outlined below.

R_2\rightarrow R_2-R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]

R_3\rightarrow R_3-2R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]

R_3\rightarrow R_3-R_2

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]

The last row determines, if there are solutions or not. To be consistent, we must have k such that

k^2-k-2=0

\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2

Case k = −1:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]

This gives the infinite many solution.

5 0
3 years ago
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kvv77 [185]

Answer:

1

Step-by-step explanation:

8-7= 1

Hope this helps!

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3 years ago
Find the perimeter. Simplify the answer.
Ugo [173]

Answer: 26t - 8

Step-by-step explanation:

7 0
3 years ago
What multiplication expression can you use to solve 2 divided by 8 4/5?
valina [46]
2*(5/44)

Dividing fractions means take the reciprocal of the second fraction then multiply
Find the reciprocal of 8 4/5
44/5 becomes 5/44
Then multiply by 2
8 0
3 years ago
On a 16 scale drawing of a bike, one part is 3 inches long. How long will the actual bike part be?
Oksanka [162]
Multiply 16*3 to get 48, the bike is 48 units so 16:1,16/1=16. So the answer would be 16. Hope I helped! 
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3 years ago
Read 2 more answers
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