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Yanka [14]
3 years ago
8

What is the quotient StartFraction 15 p Superscript negative 4 Baseline q Superscript negative 6 Baseline Over negative 20 p Sup

erscript negative 12 Baseline q Superscript negative 3 Baseline EndFraction in simplified form? Assume p not-equals 0, q not-equals 0.
answer
Negative StartFraction 3 p Superscript 8 Baseline Over 4 q cubed EndFraction
Negative StartFraction 3 Over 4 p Superscript 16 Baseline q Superscript 9 Baseline EndFraction
Negative StartFraction p Superscript 8 Baseline Over 5 q cubed EndFraction
Negative StartFraction 1 Over 5 p Superscript 16 Baseline q Superscript 9 Baseline EndFraction
Mathematics
2 answers:
svetlana [45]3 years ago
5 0

Answer:

-9

Step-by-step explanation:

I took the test.

Airida [17]3 years ago
4 0

Answer:

Answer is A (-3p^8/4q^3)

Step-by-step explanation:

You might be interested in
Which decimal number is equivalent to 1/10 ?<br> A.1<br> B.0.1<br> C.0.01<br> D.0.001
elena-14-01-66 [18.8K]
If you would like to find the decimal number which is equivalent to 1/10, you can do this using the following step:

1/10 = 0.1

The correct result would be B. 0.1.
4 0
3 years ago
Please Complete the page.
Alexxandr [17]
7. -8/3
8. 4/35
9. -21/50
10. 35/64
11.-25/54
6 0
3 years ago
Find the area **Number Only** Thanks
Contact [7]

Answer:

a=12

Step-by-step explanation:

A=b*h/2

a=4.8mi*5mi/2

a=24/2

a=12

hope this helps

7 0
3 years ago
Hey if you see this i really need your help its an exam thanks
professor190 [17]

Answer:

I think it is Letter B

Because I add the two

L-2, +N 12=10

7 0
2 years ago
With full explanation from the internet like before<br> 3x2-6x+5=0
Elodia [21]

Answer:

\sf x=1+i\sqrt{\dfrac{2}{3}}   \ \quad and  \quad \:x=1-i\sqrt{\dfrac{2}{3}}

Explanation:

<u>Given Expression</u>:

  • 3x² - 6x + 5 = 0

Use the Quadratic Formula:

\sf x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}  \  \ when   \ \  ax^2 + bx + c = 0

<u>insert coefficients</u>

\Longrightarrow \sf x = \dfrac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:3\cdot \:5}}{2\cdot \:3}

\Longrightarrow \sf x = \dfrac{\left6\right\pm \sqrt{-24} }{6}

\Longrightarrow \sf x = \dfrac{\left6\right\pm 2\sqrt{6}i}{6}

\Longrightarrow \sf x =1 \pm   i\dfrac{\sqrt{6} }{3}

\Longrightarrow \sf x=1+i\sqrt{\dfrac{2}{3}},  \quad 1-i\sqrt{\dfrac{2}{3}}

6 0
2 years ago
Read 2 more answers
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