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Orlov [11]
3 years ago
15

Point k is graphed on a number line at 3. Point l is 11 units away from point k on the number line. What could be the coordinate

s of point l? Choose all answers that are correct.
a. –14
b. –8
c. 7
d. 14
Mathematics
1 answer:
dangina [55]3 years ago
4 0

d 14 and b - 8

There are 2 possible positions for point I

11 units to the right or 11 units to the left of K

11 right gives I = 3 + 11 = 14 ⇒ d

11 left gives I = 3 - 11 = - 8 ⇒ b


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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
If 3(x − 2) = −7, then x =?
Katyanochek1 [597]

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is 3(x-2) = -7 then it's same as 3x-6=-7

We simplify this by adding 6 and subtracting 6.

Our result is

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andriy [413]

Answer:

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Step-by-step explanation:

When a quantity "A" varies DIRECTLY with another quantity "B", we can write the proportionality equation as:

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In this problem, it is given x varies DIRECTLY with y, so we can write:

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Also, given, x = 11, y = -3, so we plug these into the equation and first find k:

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Now, we need to find x when y = 8, so we substitute and find x:

x=ky\\x=(-\frac{11}{3})(8)\\x=-\frac{88}{3}

8 0
3 years ago
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