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Aleonysh [2.5K]
3 years ago
15

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!

Mathematics
1 answer:
Gekata [30.6K]3 years ago
7 0

Answer:

Its the first option

Step-by-step explanation:

Factor the denominator:-

q^2 - 7q - 8 = (q - 8)(q + 1)

The denominator must be > 0 so the restriction is q ≠ -1 and q ≠ 8.

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At 7 A.M. the temperature was -2°C. In the afternoon, the temperature was 11°C. What was the change of temperature during the da
stellarik [79]
The change was 13 degrees celsius 
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3 years ago
Use the quadratic formula to solve the equation.<br><br>-3x^2 - x - 3 = 0
vredina [299]

Answer:  -1 ± <em>i</em>√35

                  ⁻⁻⁻⁻⁻⁻⁻⁻

                     6

Step-by-step explanation:

From the quadratic equation,

X =  -b ±√b² - 4ac

     ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

             2a

From the given equation, -3x² - x - 3 = 0

where a = -3, b = -1, c = -3

Now substitute for the values in the formula above

x = -(-1) ± √(-1)² - 4(-3)(-3)

      ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

                   2(-3)

x = 1 ± √ 1 - 36

     ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

           -6

x = 1 ±  √ -35

     ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

           -6

x = -1  ±  <em>i√35   this is because it has a negative root.</em>

<em>       ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻</em>

<em>            6</em>

<em> The answer is b</em>

   

8 0
3 years ago
A salesperson sold a car for $18,200 and their commission is $700. What percentage of the sale price is their commission?
marshall27 [118]
The percentage of their commission is 3.85%
700/18200= 0.03846
0.03846 x 100 = about 3.85
7 0
3 years ago
HELP PLEASE AND THX<br> What's the slope
lara [203]

Answer:

what's the slope of what

Step-by-step explanation:

Send a pic or something.

4 0
3 years ago
Read 2 more answers
If the temperature after 4.4 degree drop was -2.5 degrees Celsius what was the temperature at noon
Lostsunrise [7]

Answer:

1.9°C

Step-by-step explanation:

Let X be the temperature at noon and -2.5°C be the current temperature after the drop.

-We calculate temperature at noon as:

\bigtriangleup =t_o-t_n\\4.4=X--2.5\\\\X=4.4-2.5\\\\=1.5\textdegree C

Hence, the temperature at noon was 1.9°C

5 0
3 years ago
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