Hi!
This is a fun one, as it delves into basic trigonometry.
We're going to use the Pythagorean theorem here, which says that for right triangles where "c" is the hypotenuse,
a² + b² = c²
We have to split this large triangle into two parts, both of which are right triangles. (This is why they drew a line in the middle to tell you that the larger triangle is composed of two right triangles.)
Let's do the one on the right first.
We know that the length of the hypotenuse is 10, and that the length of one of the legs is 6.5. If we plug this into our equation, we'll get the length of the other leg. I'm choosing "b" to be 6.5, but it really doesn't matter if you pick "a" or "b", so long as you reserve "c" for the hypotenuse (longest side).
a² + 6.5² = 10²
a² + 42.25 = 100
a² = 57.75
√a² = √57.75
a ≈ 7.6
Therefore, the length of DC is about 7.6.
Find the length of AD using the same method (7.5 is the hypotenuse "c", and 6.5 is one of the legs "a" or "b"). Then, once you have AD, add the lengths of AD and DC to get AC.
Have a great one!
Answer:
Step-by-step explanation:
{ - 9, - 4, - 1}
The answer should be b
The other one should be 8
Answer:
![f(x)=4\sqrt[3]{16}^{2x}](https://tex.z-dn.net/?f=f%28x%29%3D4%5Csqrt%5B3%5D%7B16%7D%5E%7B2x%7D)
Step-by-step explanation:
We believe you're wanting to find a function with an equivalent base of ...
![4\sqrt[3]{4}\approx 6.3496](https://tex.z-dn.net/?f=4%5Csqrt%5B3%5D%7B4%7D%5Capprox%206.3496)
The functions you're looking at seem to be ...
![f(x)=2\sqrt[3]{16}^x\approx 2\cdot2.5198^x\\\\f(x)=2\sqrt[3]{64}^x=2\cdot 4^x\\\\f(x)=4\sqrt[3]{16}^{2x}\approx 4\cdot 6.3496^x\ \leftarrow\text{ this one}\\\\f(x)=4\sqrt[3]{64}^{2x}=4\cdot 16^x](https://tex.z-dn.net/?f=f%28x%29%3D2%5Csqrt%5B3%5D%7B16%7D%5Ex%5Capprox%202%5Ccdot2.5198%5Ex%5C%5C%5C%5Cf%28x%29%3D2%5Csqrt%5B3%5D%7B64%7D%5Ex%3D2%5Ccdot%204%5Ex%5C%5C%5C%5Cf%28x%29%3D4%5Csqrt%5B3%5D%7B16%7D%5E%7B2x%7D%5Capprox%204%5Ccdot%206.3496%5Ex%5C%20%5Cleftarrow%5Ctext%7B%20this%20one%7D%5C%5C%5C%5Cf%28x%29%3D4%5Csqrt%5B3%5D%7B64%7D%5E%7B2x%7D%3D4%5Ccdot%2016%5Ex)
The third choice seems to be the one you're looking for.
20b + 18s < = 150 (thats less then or equal)
