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adell [148]
2 years ago
6

After John worked at a job for 10 years, his salary doubled. If he started at $x, his salary after 10 years is _____

Mathematics
2 answers:
Mkey [24]2 years ago
8 0

Answer:

After ten years his salary is x2

Step-by-step explanation:

erik [133]2 years ago
8 0

John’s salary after doubling in 10 years is $ 2x

<u>Solution:</u>

It is given to us that John worked at a job for 10 years.

He started working at the job with a salary of $x.

After 10 years the salary has been doubled.

We have been asked to find John’s salary now after doubling.

To find John’s salary after 10 years we need to do the following.

Salary after 10 years = Salary before 10 years \times 2

x \times 2 = 2x

Therefore, John’s salary after doubling in 10 years is $2x

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What do you mean? I don't think it's enough information
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3 years ago
F(x) = x2+3x+2 x2+5x+4 Why is there no zero at x = –1?
dmitriy555 [2]
You could rewrite F(x) as

\dfrac{x^2+3x+2}{x^2+5x+4}=\dfrac{(x+1)(x+2)}{(x+1)(x+4)}

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When x=-1, you end up with the indeterminate form \dfrac00, which is why -1 is not a zero.
7 0
3 years ago
In a triangle the measure of the first angle is twice the measure of the second angle. the measure of the third angle is 8 degre
Rufina [12.5K]
Let's call the missing angles Q, R, and S. We know that Q = 2R and S = R - 8 and we also know that Q + R + S = 180. So that really just means that              180 = 4R - 8  Add 8 on both sides of the equal sign
188 = 4R So we divide on both sides of the equal sign by 4
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The missing angles are 39 degrees 94 degrees and 47 degrees.

4 0
3 years ago
Help please I need help​
Llana [10]

Answer:

#4 -  y= 1/2x-2                        #5-  y=1/3x+3

x                    y                       x                    y

0                   -2                      0                   3

2                   -1                       3                    4

4                   0                       6                    5

6                    1                       9                    6

Step-by-step explanation:

Hope that helped :)

5 0
2 years ago
F $1 \le a \le 10$ and $1 \le b \le 36$, for how many ordered pairs of integers $(a, b)$ is $\sqrt{a + \sqrt{b}}$ an integer?
svetlana [45]
You have such entry data: 
1\ \textless \ a\ \textless \ 10 \\ 1\ \textless \ b\ \textless \ 36

Consider expression \sqrt{a+ \sqrt{b} }. If this expression becomes an integer, then b=4,9,16,25, because then \sqrt{b} = 2,3,4,5, respectively. In other cases \sqrt{b} is not integer and thus the expression \sqrt{a+ \sqrt{b} } also is not integer.

1. b=4, then \sqrt{a+ \sqrt{b} }=\sqrt{a+2}. Here a=2,7 (in other cases \sqrt{a+2} is not integer). When a=2, \sqrt{a+2}=\sqrt{2+2}=2 and when a=7, \sqrt{a+2}=\sqrt{7+2}=3.

2. b=9, then a=6 and \sqrt{a+ \sqrt{b} }= \sqrt{a+3}=\sqrt{6+3}=3.

3. b=16, then a=5 and \sqrt{a+ \sqrt{b} }= \sqrt{a+4}=\sqrt{5+4}=3.

4. b=25, then a=4 and \sqrt{a+ \sqrt{b} }=\sqrt{a+5}=\sqrt{4+5}=3.

Answer: (2,4), (7,4), (6,9), (5,16), (4,25).







3 0
3 years ago
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