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almond37 [142]
2 years ago
13

What is the domain of the relation?

Mathematics
1 answer:
White raven [17]2 years ago
8 0
The correct answer should be D
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Quadrilateral ABCD with vertices A(-9, 2), B(-8, 8), C(-4, 6), and D(-2, 2) is shown. You want to reflect quadrilateral ABCD in
telo118 [61]

Answer:

See explanation.

Step-by-step explanation:

Graph the quadrilateral. By inspection, you can tell where the midpoint is. (see attachment 1)

Now, I'll draw a line through it. (see attachment 2)

To reflect across a line, think about the points traveling across the line the same number of spaces the point is from the line. For example, point A is three away from the line. So, A' will be 3 away from the other side. The coordinates will be at (-3,2). (see attachment 3).

Do the same for the other points, and you'll have your image.

7 0
2 years ago
Maria puts $2,500.00 in her money market account earning 8% interest annually. How much interest and how much money would she ha
Gnom [1K]
The correct answer: total interest = $1,784.56 and she would have $4,284.56 after 7 years
7 0
2 years ago
Lf g(x) = x3 - 5 and h(x) = 2x - 2, find g(h(3)).
White raven [17]

Answer: g(h(3)) = 59

Step-by-step explanation:

Well if its x^3 I assume it is because you wrote other coefficients before the variable.

2(3) - 2 = 6-2 = 4

So 4 is the x input for h(x)

4^3 -5 = 64-5 = 59

7 0
3 years ago
For what values of θ on the polar curve r=θ, with 0≤θ≤2π , are the tangent lines horizontal? Vertical?
Bond [772]
Given that r=\theta, then r'=1

The slope of a tangent line in the polar coordinate is given by:

m= \frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}

Thus, we have:

m= \frac{\sin\theta+\theta\cos\theta}{\cos\theta-\theta\sin\theta}



Part A:

For horizontal tangent lines, m = 0.

Thus, we have:

\sin\theta+\theta\cos\theta=0 \\  \\ \theta\cos\theta=-\sin\theta \\  \\ \theta=- \frac{\sin\theta}{\cos\theta} =-\tan\theta

Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:

</span><span>θ = 0

</span>θ = <span>2.02875783811043
</span>
θ = <span>4.91318043943488



Part B:

For vertical tangent lines, \frac{1}{m} =0

Thus, we have:

\cos\theta-\theta\sin\theta=0 \\  \\ \Rightarrow\theta\sin\theta=\cos\theta \\  \\ \Rightarrow\theta= \frac{\cos\theta}{\sin\theta} =\sec\theta

</span>Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:

</span>θ = <span>4.91718592528713</span>
3 0
3 years ago
Please help me out on this question
Mariulka [41]

At x=2, y=4

At x=4, y=16

So the total change in y from x=2 to x=4 is 16-4=12

And since average rate of change = total change in y divided by total change in x within the same period, therefore the answer is 12/2 = 6

Hope that helps. Let me know if you have any questions.

7 0
3 years ago
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