Answer:
14 I think I'm. it completely sure
The solution to the given expression negative four and one third ÷ two and one fifth is -1 32/33
<h3>Fraction division</h3>
negative four and one third ÷ two and one fifth
-4 1/3 ÷ 2 1/5
= -13/3 ÷ 11/5
- Multiply by the reciprocal of 11/5
- The reciprocal of 11/5 is 5/11
= -13/3 × 5/11
= (-13 × 5) / (3 × 11)
= -65 / 33
= -1 32/33
Therefore, the solution to the fraction is -1 32/33
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Perpendicular lines have slopes that multiply to -1
get into y=mx+b form
minus 15x both sides and divide by -5
y=3x-7/5
slope is 3
3 times what=-1?
what=-1/3
so the slope is -1/3
so
y=-1/3x+b
we use the point (0,-4)
x=0 and y=-4
-4=-1/3(0)+b
-4=b
y=(-1/3)x-4 is da equation
It’s like -5 or something maybe
Answer:

Step-by-step explanation:
Given equation:

Cube root both sides:
![\implies \sqrt[3]{p^3}= \sqrt[3]{\dfrac{1}{8}}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csqrt%5B3%5D%7Bp%5E3%7D%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B8%7D%7D)
![\implies p= \sqrt[3]{\dfrac{1}{8}}](https://tex.z-dn.net/?f=%5Cimplies%20p%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B8%7D%7D)
![\textsf{Apply exponent rule} \quad \sqrt[n]{a}=a^{\frac{1}{n}}:](https://tex.z-dn.net/?f=%5Ctextsf%7BApply%20exponent%20rule%7D%20%5Cquad%20%5Csqrt%5Bn%5D%7Ba%7D%3Da%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%3A)





Rewrite 8 as 2³:



Simplify:


