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3 years ago

Please give me the right answer

Mathematics

1 answer:

ivann1987 [24]3 years ago

8 0

Answer:j

Step-by-step explanation:

If it’s closed it would be greater than -2 but since it’s open it’s equal or greater than

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Which graph represents the function f(x) = 1/3 |x|

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Lina20 [59]

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You correctly answered!

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Jamie kicked a soccer ball that traveled along a path described by the parabola y = -x^2 +8x, where y equals the height of the s

andrey2020 [161]

We have that
<span>y = -x</span>²<span>+8x

we know that
when the ball hint the ground is the x-intercepts of the graph
the x-intercepts is for y=0
so
0=-x</span>²+8x--------> x²-8x=0----> x*(x-8)=0

the solution is
x=0
x=8

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the distance between the x intercepts points =[8-0]-----> 8 ft

the answer is
8 ft

see the attached figure

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3 years ago

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Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Anna [14]

Answer:

$\large\boxed{1\dfrac{1}{3}\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}$

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3 years ago