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3 years ago

Please help i need the answer asap

Mathematics

1 answer:

Sav [38]3 years ago

3 0

i do not understand what you are asking help for?

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Income tax paid to the federal government is based on your _______________.

Kryger [21]

Answer: C final income

Step-by-step explanation: I might be wrong but im pretty sure that its C.

Good Luck!

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3 years ago

Read 2 more answers

Write inequalities to represent the situations below. The cargo of the truck weighs no more than 2,000 pounds. Use w to represen

Alex777 [14]

Answer:

w ≤ 2000 lb.

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3 years ago

People buy me to eat, but never eat me. What am I?

egoroff_w [7]

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4 years ago

Solve x/12+1/3=5/6 please help me resolve

viktelen [127]

Subtract 1/3 from both sides

x/12 = 5/6 - 1/3

Simplify 5/6 - 1/3 to 1/2

x/12 = 1/2

Multiply both sides by 12

x = 12 × 12

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x = 12/2

Simplify 12/2 to 6

4 0

3 years ago

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:

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

6 0

3 years ago