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lara [203]
2 years ago
11

imes \sqrt{12}" alt="\sf \sqrt{12} \times \sqrt{12}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Darina [25.2K]2 years ago
7 0

Answer:

12

Step-by-step explanation:

\sqrt{12} \times\sqrt{12}

\mathrm{Apply\:radical\:rule}:\quad \sqrt{a}\sqrt{a}=a,\:\quad \:a\ge 0

\sqrt{12}\sqrt{12}=12

=12

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Need some help working these out
s2008m [1.1K]

Answers:

  • Problem 1)  40 degrees
  • Problem 2) 84 degrees
  • Problem 3) 110 degrees

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

Explanation:

For these questions, we'll use the inscribed angle theorem. This says that the inscribed angle is half the measure of the arc it cuts off. An inscribed angle is one where the vertex of the angle lies on the circle, as problem 1 indicates.

For problem 1, the arc measure is 80 degrees, so half that is 40. This is the measure of the unknown inscribed angle.

Problem 2 will have us work in reverse to double the inscribed angle 42 to get 84.

-------------------

For problem 3, we need to determine angle DEP. But first, we'll need Thales Theorem which is a special case of the inscribed angle theorem. This theorem states that if you have a semicircle, then any inscribed angle will always be 90 degrees. This is a handy way to form 90 degree angles if all you have is a compass and straightedge.

This all means that angle DEF is a right angle and 90 degrees.

So,

(angle DEP) + (angle PEF) = angle DEF

(angle DEP) + (35) = 90

angle DEP = 90 - 35

angle DEP = 55

The inscribed angle DEP cuts off the arc we want to find. Using the inscribed angle theorem, we double 55 to get 110 which is the measure of minor arc FD.

3 0
3 years ago
Please help functions
katrin2010 [14]

A function is an equation or graph where each x value only has one y value. In the first image you can use the vertical line test. Take your pencil and lay it on the paper vertically. Move it across the graph. Basically if an x value has multiple y value it is not a function. The first image is not a function because all x values past zero have multiple y values for each x value. In the second image it is not A because in A the x value of 2 has a y value of 1 and 8. It is not B because the 5 has a y value of 3 and 7. It’s is not D because of the 5 again. It is C because all x values are unique.

Hope that helps

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3 years ago
Could u answer all and explain it to me?
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I got the same book with those same answers ima bout to get it ok

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3 years ago
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3 years ago
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11/3....this is rational because it is a fraction
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4 0
3 years ago
Read 2 more answers
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