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kow [346]
3 years ago
7

Each classmate contributes $2 for charity. Write an expression for the amount of raised by your class.

Mathematics
1 answer:
Semenov [28]3 years ago
6 0
(amount raised) = $2 × (number of classmates)
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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}

6 0
3 years ago
Is 0.06 greater than 0.0582
Art [367]

Yes, 0.06 is greater than 0.0582.

If we look at the hundreths place, we can see 6 and 5. 6 is greater than 5 which proves that 0.06 is greater.

Best of Luck!

5 0
3 years ago
Ay anyone willing to help me with my work kinda lost
kirill115 [55]

Okaayyy

Set B has the greater mode

7 0
3 years ago
Given that ∠S ≅ ∠A, and AB ≅ ST , what is the third congruence needed to prove that ΔABC ≅ ΔSTU by ASA?
Olenka [21]

Answer:

angle-side-angle. that's my answer

7 0
3 years ago
When n=1 there is 1 dot. when n=2 there are 3 dots, when n=3 there are 6 dots. notice that the total number of dots increases by
gayaneshka [121]

Answer:

<h3>Part A</h3>
  • d(n) = n(n + 1)/2
  • n = 1 ⇒ d(1) = 1(1 + 1)/2 = 2/2 = 1, proved
<h3>Part B</h3>
  • d(k) + (k + 1) =
  • k(k + 1)/2 + (k + 1) =
  • [k(k + 1) + 2(k + 1)]/2 =
  • [(k + 1)(k + 1 + 1)]/2 =
  • d(k + 1)
  • Proved

6 0
2 years ago
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