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4vir4ik [10]
4 years ago
15

I need help with question a and b

Mathematics
1 answer:
DaniilM [7]4 years ago
5 0

Solving for part A

f(x) = x^{2} \;\;and\;\; g(x) = \frac{1}{x^{3}} \\\\f(g(x)) = (\frac{1}{x^{3}})^{2}   =\frac{1}{x^{6}}

Domain of composition: x ≠ 0 i.e. Domain is the set of "Real numbers excluding number 0".

Solving for part B

f(x) = \frac{x}{x-2}\;\;and\;\;g(x) =\frac{3}{x} \\\\f(g(x))=\frac{\frac{3}{x}}{\frac{3}{x}-2}=\frac{\frac{3}{x}}{\frac{3-2x}{x}} =\frac{3}{3-2x}\\\\  Domain:\;3-2x\neq  0 \;\;OR\;\;x\neq \frac{3}{2}

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In 30 minutes the minute hand makes half a turn around the clock, so it sweeps half a circle.

Given the radius r, the area of a circle is A = \pi r^2

In this case, we're interested in half the area of a circle with radius 14cm, so we have

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Mashutka [201]

Answer:

we choose (3,4)

Step-by-step explanation:

Given the coordinates:

  • A (5, 2)
  • B (5, -2)
  • C  (2,1)
  • L (-5, 6)
  • M (-5, -6)

To find a distance between two points or the length of the segment, we use the following formula:

\sqrt{(x2-x1)^2+(y2-y1)^2}

  • Because A and B are located in the same line x =5

=> the lenght of AB = \sqrt{(-2-2)^{2} } =\sqrt{-4^{2} } =4

  • Because A and B are located in the same line x =-5

=>  the lenght LM = \sqrt{(-6-6)^{2} } =\sqrt{-12^{2} } =12

  • Because ABC is similar to LMN

=> the ratio of LM : AB = 12:4 = 3:1

<=> the ratio LN : CA = 3:1

We need to find the lenght of CA

= \sqrt{(2-5)^2+(1-2)^2} = \sqrt{-3^{2} + -1^{2} } = \sqrt{10}

=> the lenght of LN = 3CA = 3\sqrt{10}

Let (x, y) is the coordinate of N, we have:

The length of LN: \sqrt{(x+5)^2+(y-6)^2} = 3\sqrt{10} = \sqrt{90}

Let try all the possible answer;

(3,4) => \sqrt{(3+5)^2+(4-6)^2} = \sqrt{8^{2} + -2^{2} } = \sqrt{90} True

So we choose (3,4)

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I think it will be rectangle or square

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Each of the line segments in the word MATH are numbered in the graph below. Find the slope (as a ratio of rise over run) of each
Leviafan [203]

Answer:

Line 1: \infty

Line 2: -\frac{4}{3}

Line 3: \frac{4}{3}

Line 4: \infty

Line 5: \frac{7}{4}

Line 6 : -\frac{7}{3}

Line 7: 0

Line 8: 0

Line 9: \infty

Line 10: \infty

Line 11: 0

Line 12: \infty

Step-by-step explanation:

Slope of line is given by the formula:

\text{Slope = }\dfrac{\text{Change in } y\text{ coordinate}}{\text{Change in } x\text{ coordinate}}

Line 1:

The change in x coordinate is zero.

Therefore slope of line 1 is \infty.

Line 2:

Change in y coordinate = -4

Change in x coordinate = 3

Slope = -\frac{4}{3}

Line 3:

Change in y coordinate = 4

Change in x coordinate = 3

Slope = \frac{4}{3}

Line 4:

The change in x coordinate is zero.

Therefore slope of line 4 is \infty.

Line 5:

Change in y coordinate = 7

Change in x coordinate = 3

Slope = \frac{7}{3}

Line 6:

Change in y coordinate = -7

Change in x coordinate = 3

Slope = -\frac{7}{3}

Line 7:

Change in y coordinate = 0

Slope = 0

Line 8:

Change in y coordinate = 0

Slope = 0

Line 9:

The change in x coordinate is zero.

Therefore slope of line 9 is \infty.

Line 10:

The change in x coordinate is zero.

Therefore slope of line 10 is \infty.

Line 11:

The change in y coordinate is zero.

Therefore slope of line 11 is 0.

Line 12:

The change in x coordinate is zero.

Therefore slope of line 12 is \infty.

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