f(h(x))= 2x -21
Step-by-step explanation:
f(x)= x^3 - 6
h(x)=\sqrt[3]{2x-15}
WE need to find f(h(x)), use composition of functions
Plug in h(x)
f(h(x))=f(\sqrt[3]{2x-15})
Now we plug in f(x) in f(x)
f(h(x))=f(\sqrt[3]{2x-15})=(\sqrt[3]{2x-15})^3 - 6
cube and cube root will get cancelled
f(h(x))= 2x-15 -6= 2 x-21
wait i think i did the wrong one brb
Problem 1
<h3>Answer: 7.3</h3>
Explanation: Apply the square root to the area to get the side length. This only applies to areas that are squares (hence the name).
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Problem 2
<h3>Answer: C) 1.3</h3>
Explanation: Use your calculator to find that choices A,B,D plugged into the square root function yield terminating decimal values. "Terminating" means "stop". This implies that they are perfect squares (though not perfect squares in the sense of whole number perfect squares which you may be used to). Choice C is the only value that has a square root that leads to a non-terminating decimal. The digits of this decimal go on forever without any pattern. The value is irrational.
- sqrt(5.29) = 2.3 terminating decimal
- sqrt(13.69) = 3.7 terminating decimal
- sqrt(1.3) = 1.140175425 keeps going forever without any pattern
- sqrt(0.09) = 0.3 terminating decimal
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Problem 3
<h3>Answer: 23.6 feet approximately</h3>
Explanation: Apply the square root to 15.5 to get roughly 3.937; this is the approximate side length of one square. Six of these tiles placed together will lead to a total length of roughly 6*3.937 = 23.622 which rounds to 23.6 feet. Like with problem 1, the square root being used like this only works for square areas.
Dang that’s crazy, send me some pics and you got a deal
Choices A, C, D have a y-value that is 5 more than the x-value, so those points are on the line defined by the first equation.
Choice D satisfies the requirement
... y = 2x - 13
... 23 = 2·18 - 13
... 23 = 36 - 13
The appropriate choice is ...
... D (18, 23)
