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3 years ago

Which of the sets of ordered pair represents a function?

Mathematics

1 answer:

Ratling [72]3 years ago

3 0

Answer:

both p and q

Step-by-step explanation:

This is because each x coordinate is mapped to a member in the co domain

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A pair of pants with a marked tag of $35.00 has a discount of 20%. How much is the discount?

melisa1 [442]

$35.00 x .20 = $7

So the discount would be $7 off of $35, for a final sale price of $28

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3 years ago

Sophie [7]

Answer:

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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:

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

6 0

3 years ago

Find the least common multiple for 6(x+1)^3(x-4)^2 and 10(x+1)^8(x-4)^5

dolphi86 [110]

Answer:

30 $(x+1)^8 (x-4)^5$

Step-by-step explanation

In order to find the Least common multiple we write each of the factor only once from each of the expression and for the common expression we take their LCM as maximum of the exponent in all expressions

as here in the question exponent of (x+1) are 3 and 8 so we take exponent 8

likewise for (x-4) we shall take maximum of 2 and 5 which is 5

so our expression for Least common multiple will be

2X3X5 X $(x+1)^8 (x-4)^5$

30 $(x+1)^8 (x-4)^5$

7 0

3 years ago

Given: F(x) = 2x - 1; G(x) = 3x + 2; H(x) = x 2 Find F[G(x)] - F(x). 4x + 4 4x + 2 4x

Allushta [10]

Answer:

4x + 4

Step-by-step explanation:

F(x) = 2x - 1

G(x) = 3x + 2

H(x) = x²

We have to calculate the expression F(G(x)) - F(x)

F(G(x)) means the composition of functions F(x) and G(x). In order to find F(G(x)) we have to replace ever occurrence of x in F(x) with the value of G(x). So,

F(G(x)) = 2(3x+2) - 1 = 6x + 4 - 1 = 6x + 3

Thus,

F(G(x)) - F(x) = 6x + 3 - (2x - 1)

= 6x + 3 -2x + 1

= 4x + 4

Therefore, the expression F(G(x)) - F(x) equals 4x + 4

4 0

3 years ago