We'll use standard labeling of right triangle ABC, C=90 degrees, legs a, b, hypotenuse c.
11.
Right triangle, cliff peak A, boat B, angle opposite cliff is B=28.9 deg. adjacent leg a=65.7 m, cliff height is leg b.
tan B = b/a
b = a tan B = 65.7 tan 28.9° = 36.3 m
12.
Similar story, boat at B, opposite b=3.5 m, rope c=12 m
sin B = b/c
B = arcsin b/c = arcsin (3.5/12) = 17.0°
13.
c=124 m, A=58°
sin A = a/c
a = c sin A = 124 sin 58 = 105.2 m
14.
That's a hypotenuse c=4-1.2 = 2.8 m to a height b=1.8m so
cos A = b/c
A = arccos b/c = arccos (1.8/2.8) = 50.0°
15.
Not a right triangle, an isosceles triangle. Half of it is a right triangle with hypotenuse one arm, c=9.8 cm and angle opposite half the base of B=62/2=31°. We're after d=2b:
sin B = b/c
b = c sin B
d = 2b = 2 c sin B = 2(9.8) sin 31 = 10.1 cm
Almost equilateral
k<2(3.5) or simplified k<7
I dont know if this is what you are looking for but i think this is the answer of the question you are looking for which is the general equation of a horizontal hyperbola:
<span>(x-h)²/a² - (y-k)²/b² = 1 </span>
<span>with </span>
<span>center (h,k)
</span>What you need to do is replace the data and do the calculations
Answer:
![x^\frac{21}{5}](https://tex.z-dn.net/?f=x%5E%5Cfrac%7B21%7D%7B5%7D)
Step-by-step explanation:
We are given an expression and we have to transform it into an expression with exponent.
5th root of x can be written as ![x^\frac{1}{5}](https://tex.z-dn.net/?f=x%5E%5Cfrac%7B1%7D%7B5%7D)
![(\sqrt[5]{x^7})^3\\\\((x^7)^\frac{1}{5} )^3](https://tex.z-dn.net/?f=%28%5Csqrt%5B5%5D%7Bx%5E7%7D%29%5E3%5C%5C%5C%5C%28%28x%5E7%29%5E%5Cfrac%7B1%7D%7B5%7D%20%29%5E3)
The powers outside will be multiplied as: ![(x^a)^b = x^{ab}](https://tex.z-dn.net/?f=%28x%5Ea%29%5Eb%20%3D%20x%5E%7Bab%7D)
![(x^\frac{7}{5} )^3\\x^\frac{7*3}{5}\\x^\frac{21}{5}](https://tex.z-dn.net/?f=%28x%5E%5Cfrac%7B7%7D%7B5%7D%20%29%5E3%5C%5Cx%5E%5Cfrac%7B7%2A3%7D%7B5%7D%5C%5Cx%5E%5Cfrac%7B21%7D%7B5%7D)
where the exponent is
and it is a rational number by definition of rational numbers
The answer is D. -24 - 49i/13