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ZanzabumX [31]
3 years ago
5

Please help me with numbers #2-5?

Mathematics
1 answer:
trasher [3.6K]3 years ago
7 0

Just keep doing what you did in 1. I'll show you how easy it is.

2.

a) g(9) = 9 - 5 / 2 = 4 / 2 = 2.

b) g(0) = 0 - 5 / 2 = 5 / 2 = 2 1 / 2.

c) g(3) = 3 - 5 / 2 = 2 / 2 = 1.

d) g(17) = 17 - 5 / 2 = 12 / 2 = 6.

3.

a) f(3) = 3^2 - 4 = 9 - 5 = 4.

b) f(-4) = -4^2 - 4 = 16 - 4 = 12.

c) f(0) = 0^2 - 4 = 0 - 4 = -4.

d) f(-2) = -2^2 - 4 = 4 - 4 = 0.

4.

a) f(10) = 10 / 2 - 6 = 5 - 6 = -1; -1 is the solution.

5.

a) I'll test one after another.

1. f(0) = 2(0) - 3 = 0 - 3 = -3  >  g(0) = 3(0) / 2 + 1 = 0 + 1 = 1; this is incorrect.

2. f(2) = 2(2) - 3 = 4 - 3 = 1  = g(2) = 3(2) / 2 + 1 = 6 / 2 + 1 = 3 + 1 = 4; this is incorrect.

3. f(8) = 2(8) - 3 = 16 - 3 = 13 = g(8) = 3(8) / 2 + 1 = 24 / 2 + 1 = 12 + 1 = 13; this is correct.

4. g(4) = 3(4) / 2 + 1 = 12 / 2 + 1 = 6 + 1 = 7 < f(4) = 2(4) - 3 = 8 - 3 = 5; this is incorrect.

Hope this helps :)

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The graph from which the position of the point <em>A</em> can determined following

the multiplication with a scalar is attached.

Responses:

  • If <em>A</em> is in quadrant I and is multiplied by a negative scalar, <em>c</em>, then c·A is in <u>quadrant III</u>
  • If A is in quadrant II and is multiplied by a positive scalar, <em>c</em>, then c·A is in <u>quadrant II</u>
  • If <em>A</em> is in quadrant II and is multiplied by a negative scalar, <em>c</em>, then c·A is in <u>quadrant IV</u>
  • If <em>A</em> is in quadrant III and is multiplied by a negative scalar, <em>c</em>, then c·A is in <u>quadrant I</u>

<h3>Methods by which the above responses are obtained</h3>

Background information;

The question relates to the coordinate system with the abscissa represent the real number and the ordinate representing the imaginary number.

Solution:

If A is in quadrant I; A = a + b·i

When multiplied by a negative scalar, <em>c</em>, we get;

c·A = c·a + c·b·i

Therefore;

c·a is negative

c·b is negative

  • c·A = c·a + c·b·i is in the <u>quadrant III</u> (third quadrant)

If A is quadrant II, we have;

A = -a + b·i

When multiplied by a positive scalar <em>c</em>, we have;

c·A = c·(-a) + c·b·i = -c·a + c·b·i

-c·a is negative

c·b·i is positive

Therefore;

  • c·A = -c·a + c·b·i is in <u>quadrant II</u>

Multiplying <em>A</em> by negative scalar if <em>A</em> is in quadrant II, we have;

c·A = -c·a + c·b·i

-c·a is positive

c·b·i is negative

Therefore;

c·A = -c·a + c·b·i is in <u>quadrant IV</u>

If A is in quadrant III, we have;

A = a + b·i

a is negative

b is negative

Multiplying <em>A</em> with a negative scalar <em>c</em> gives;

c·A = c·a + c·b·i

c·a is positive

c·b  is positive

Therefore;

  • c·A = c·a + c·b·i is in<u> quadrant I</u>

Learn more about real and imaginary numbers here;

brainly.com/question/5082885

brainly.com/question/13573157

4 0
2 years ago
12x+24 is equivalent to 4(3x+6) because...
ANEK [815]

Answer:

you use disputation

Step-by-step explanation:

4 times 3x = 12x

4 times 6 = 24

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Kira drew PQR and STU so that P S, Q T, PR = 12, and SU = 3. Are PQR and STU similar? If so, identify the similarity postulate o
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ch4aika [34]

Answer:

-\dfrac{3}{5}

Step-by-step explanation:

1. Find the numerator:

\dfrac{1}{3}-\dfrac{5}{6}=\dfrac{1\cdot 2}{3\cdot 2}-\dfrac{5}{6}=\dfrac{2}{6}-\dfrac{5}{6}=\dfrac{2-5}{6}=-\dfrac{3}{6}=-\dfrac{1}{2}.

2. Simplify the fraction:

\dfrac{-\frac{1}{2}}{\frac{5}{6}}=-\dfrac{1}{2}\cdot \dfrac{6}{5}=-\dfrac{3}{5}.

7 0
3 years ago
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