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Svetllana [295]
3 years ago
12

Which view is not correct for the figure above?

Mathematics
1 answer:
Bezzdna [24]3 years ago
4 0

Answer:

Front

Step-by-step explanation:

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Find the domain of f(x) =√3x-1
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The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

( − ∞ , ∞ )  

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4 0
2 years ago
A soccer ball is kicked from 3 feet above the ground. the height,h, in feet of the ball after t seconds is given by the function
dexar [7]
Here's our equation.

h=-16t+64t+3

We want to find out when it returns to ground level (h = 0)

To find this out, we can plug in 0 and solve for t.

0 = -16t+64t+3 \\ 16t-64t-3=0 \\ use\ the\ quadratic\ formula\ \frac{-b\±\sqrt{b^2-4ac}}{2a}  \\ \frac{-(-64)\±\sqrt{(-64)^2-4(16)(-3)}}{2*16}  = \frac{64\±\sqrt{4096+192}}{32}

= \frac{64\±\sqrt{4288}}{32} = \frac{64\±8\sqrt{67}}{32} = \frac{8\±\sqrt{67}}{4} = \boxed{\frac{8+\sqrt{67}}{4}\ or\ 2-\frac{\sqrt{67}}{4}}

So the ball will return to the ground at the positive value of \boxed{\frac{8+\sqrt{67}}{4}} seconds.

What about the vertex? Simple! Since all parabolas are symmetrical, we can just take the average between our two answers from above to find t at the vertex and then plug it in to find h!

\frac{1}2(\frac{8+\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(2+\frac{\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(4) = 2

h=-16t^2+64t+3 \\ h=-16(2)^2+64(2)+3 \\ \boxed{h=67}


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3 years ago
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Step-by-step explanation:

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2 years ago
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