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k0ka [10]
3 years ago
12

Brad the pair of fractions as a pair of fractions with a common denominator 3/10 and 1/2​

Mathematics
1 answer:
LuckyWell [14K]3 years ago
4 0

Answer:

<em><u>2/10 and 3/10</u></em>

Step-by-step explanation:

Hello!

The best way of finding common denominators is using their factors!

(of denominators)

10 : 1, 2, 5, 10

2 : 1, 2

We also can see that 10 is divisible by 2.

For that reason, 1/2 is equivalent to 2 * 10 ÷ 2.

That's 10.

So <em>5/10 and 3/10</em> have a common denominator 10.

Hope this helps!

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Question 11

The directrix is a horizontal line, which means the parabola opens either upward or downward. In this case, it opens downward. This is because all answer choices have a negative leading coefficient. Also, it's because the focus is below the directrix.

For vertically opening parabolas, we use this form

4p(y-k) = (x-h)^2

where (h,k) is the vertex and p is the focal distance, aka the distance from the vertex the focus. To find (h,k), we start at the focus (0,-4) and move directly up until we reach the directrix y = 4. We'll arrive at (0,4). The midpoint of (0,-4) and (0,4) is (0,0) which is the vertex's location. So (h,k) = (0,0).

Note that in moving from (0,-4) to (0,4) is a span of 4 units. So this is the value of p.

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4p(y-k) = (x-h)^2

4*4(y-0) = (x-0)^2

16y = x^2

y = (1/16)x^2

The only adjustment we need to make is to change the 1/16 to -1/16 so that the parabola opens downward.

<h3>Answer:  Choice D.  y = -(1/16)x^2</h3>

===============================================

Question 3

The given equation is in the form y = ax^2+bx+c

In this case,

  • a = 2
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  • c = 3

Let's compute the x coordinate of the vertex h

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h = -4/(2*2)

h = -1

This h value is plugged into the original function to find k

f(x) = 2x^2+4x+3

f(-1) = 2(-1)^2+4(-1)+3

f(-1) = 1

We find that h = -1 and k = 1 pair up together. In short, (h,k) = (-1,1) is the vertex.

<h3>Answer: Choice B.  (-1,1)</h3>
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Answer:

  1. D. f^-1(x) = log2(x -6)
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1. When you replace f(x) by x and x by y, you have

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The first thing you do is subtract 6; then you take the base-2 logarithm:

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... log2(x -6) = y = f^-1(x)

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2. When you swap x and y and solve for y, you have ...

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... (-2x)^2 = y +3 . . . . . square

... 4x^2 - 3 = y = f^-1(x) . . . . subtract 3

The range of f(x) is (-∞, 0], so that is the domain of f^-1(x). That is, f^-1(x) is defined for x ≤ 0.

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