Answer:
![\large\boxed{\ln\sqrt[3]{e^4}=\dfrac{4}{3}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B%5Cln%5Csqrt%5B3%5D%7Be%5E4%7D%3D%5Cdfrac%7B4%7D%7B3%7D%7D)
Step-by-step explanation:
![\text{Use}\\\\\sqrt[n]{a^m}=a^\frac{m}{n}\\\\\ln a^n=n\ln a\\\\\ln e=1\\-----------\\\\\ln\sqrt[3]{e^4}=\ln e^\frac{4}{3}=\dfrac{4}{3}\ln e=\dfrac{4}{3}\cdot1=\dfrac{4}{3}](https://tex.z-dn.net/?f=%5Ctext%7BUse%7D%5C%5C%5C%5C%5Csqrt%5Bn%5D%7Ba%5Em%7D%3Da%5E%5Cfrac%7Bm%7D%7Bn%7D%5C%5C%5C%5C%5Cln%20a%5En%3Dn%5Cln%20a%5C%5C%5C%5C%5Cln%20e%3D1%5C%5C-----------%5C%5C%5C%5C%5Cln%5Csqrt%5B3%5D%7Be%5E4%7D%3D%5Cln%20e%5E%5Cfrac%7B4%7D%7B3%7D%3D%5Cdfrac%7B4%7D%7B3%7D%5Cln%20e%3D%5Cdfrac%7B4%7D%7B3%7D%5Ccdot1%3D%5Cdfrac%7B4%7D%7B3%7D)
Answer:
P ≈ 48.89°(nearest hundredth)
Step-by-step explanation:
The triangle PQR forms a right angle triangle since angle R is 90°. The triangle has an hypotenuse , adjacent and opposite side.
Using the SOHCAHTOA principle one can find the sine ratio of angle P. Let us designate where each side represent.
opposite side(QR) = 55
adjacent side(PR) = 48
hypotenuse(PQ) = 73
sin P = opposite/hypotenuse
sin P = 55/73
P = sin⁻¹ 55/73
P = sin⁻¹ 0.75342465753
P = 48.8879095605
P ≈ 48.89°(nearest hundredth)
Find two points on the graph that the line crosses through almost perfectly. It looks like (1,10) and (9,1) will do.
Use them to compute the slope:
m = (1 - 10) / (9 - 1)
= -9/8
Then set up the "point-slope form":
y - y0 = m * (x - x0)
You choose some point (x0, y0) that the line crosses through. We already know the line passes through (1,10) pretty well, so let's use that.
x0 = 1
y0 = 10
Now finish plugging into the equation:
y - 10 = -9/8 * (x - 1)
The above equation will work fine for an answer, but let's go a step further and solve for y.
y - 10 = -9/8x + 9/8
y = -9/8x + 9/8 + 10
y = -9/8x + 9/8 + 80/8
y = -9/8x + (9 + 80)/8
y = -9/8x + 89/8
Step-by-step explanation:
I think B or F is the best answer
Answer:
-2
Step-by-step explanation:
