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amid [387]
3 years ago
12

Please help I don’t understand.

Mathematics
1 answer:
Lina20 [59]3 years ago
4 0
You can work out c first. That's probably the key to the whole problem.
The adjacent side to a 60o angle is 1/2 the hypotenuse
The hypotenuse in this case = 4 Sqrt(3)
Then c = 1/2 (4 sqrt(3)) 
c = 2 sqrt(3) That means d is not true.

Next work out a.
a is in the same triangle as c and the hypotenuse.
a^2 + c^2 = hypotenuse^2
a = ??
c = 2 sqrt(3)
h = 4 sqrt(3)
a^2 + (2 sqrt(3))^2 = (4 sqrt(3))^2
(sqrt(3))^2 = 3
a^2 + 4 * 3 = 16 * 3
a^2 + 12 = 48
a^2 = 48 - 12
a^2 = 36
a = 6

Now we need to work out d
The side opposite and the side adjacent  are equal when opposite a 45o angle in a right angle triangle
d = 6


The last thing to work out is be
a = 6
d = 6
c = ???

a^2 + d^2 = c^2
6^2 + 6^2 = c^2
c^2 = 72
c = sqrt(72)
c = sqrt(6*6*2)
c = 6 sqrt(2)

The answer should be B??? Check this out.
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Given three collinear points A,B,C with B between A and C, four different rays can be named using points AB, BA, BC, CB. How man
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Answer:

Given n collinear points, 2(n -1) or 2n - 2 rays can be named

Step-by-step explanation:

When we talk of collinear points, we mean points that lie on the same straight line.

For 3 collinear points we can have 4 rays

For 4 collinear points, let’s say ABCD

A B C D

The rays are AB BA BC CB CD and DC making a total of 6

For 5 collinear points,

A B C D E

The rays are;

AB BA BC CB CD DC DE ED which makes a total of 8

For 6 collinear points, we have;

A B C D E F

The rays are;

AB BA BC CB CD DC DE ED FE EF which makes a total of 10

So what pattern do we notice?

3 points 4 rays

4 points 6 rays

5 points 8 rays

6 points 10 rays

7 points 12 rays

So using the pattern

n rays = 2n - 2 rays or simply 2(n - 1) rays

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4 years ago
Y = x^2+ 2x + 7<br> y = 7 + x
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Answer:

when x=0 y=7

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Step-by-step explanation:

to find x & y we can make the two x expressions equal to each other

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x=0 x=-1 (two solutions for x)

now we just plug in these values and find the y values

when x=0 y=7

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Match the function in the left column with its period in the right column.
kiruha [24]

The results for the matching between function and its period are:

  • Option 1 - Letter D
  • Option 2 - Letter A
  • Option 3 - Letter C
  • Option 4 - Letter B

<h3>What is a Period of a Function?</h3>

If a given function presents repetitions, you can define the period as the smallest part of this repetition. As an example of periodic functions, you have: sin(x) and cos(x).

\mathrm{Period\:of\:}a\cdot \cos \left(bx\:+\:c\right)\:+\:d=\frac{\mathrm{periodicity\:of}\:\cos \left(x\right)}{|b|}

\mathrm{Period\:of\:}a\cdot \sin \left(bx\:+\:c\right)\:+\:d=\frac{\mathrm{periodicity\:of}\:\sin \left(x\right)}{|b|}

The period of sin(x) and cos(x) is 2π.

For solving this question, you should analyze each option to find its period.

1) Option 1

\mathrm{Periodicity\:of\:}a\cdot \cos \left(bx\:+\:c\right)\:+\:d=\frac{\mathrm{periodicity\:of}\:\cos \left(x\right)}{|b|}\\ \\ \mathrm{Periodicity\:of\:}a\cdot \cos \left(bx\:+\:c\right)\:+\:d=\frac{2\pi }{\frac{1}{2} }=4\pi

Thus, the option 1 matches with the letter D.

2) Option 2

\mathrm{Period\:of\:}a\cdot \sin \left(bx\:+\:c\right)\:+\:d=\frac{\mathrm{periodicity\:of}\:\sin \left(x\right)}{|b|}\\ \\ \mathrm{Period\:of\:}a\cdot \sin \left(bx\:+\:c\right)\:+\:d=\frac{2\pi}{4} =\frac{\pi }{2}

Thus, the option 2 matches with the letter A.

3) Option 3

\mathrm{Periodicity\:of\:}a\cdot \cos \left(bx\:+\:c\right)\:+\:d=\frac{\mathrm{periodicity\:of}\:\cos \left(x\right)}{|b|}\\ \\ \mathrm{Periodicity\:of\:}a\cdot \cos \left(bx\:+\:c\right)\:+\:d=\frac{2\pi}{2} =\pi

Thus, the option 3 matches with the letter C.

4) Option 4

\mathrm{Period\:of\:}a\cdot \sin \left(bx\:+\:c\right)\:+\:d=\frac{\mathrm{periodicity\:of}\:\sin \left(x\right)}{|b|}\\ \\ \mathrm{Period\:of\:}a\cdot \sin \left(bx\:+\:c\right)\:+\:d=\frac{2\pi}{8} =\frac{\pi }{4}

Thus, the option 4 matches with the letter B.

Read more about the period of a trigonometric function here:

brainly.com/question/9718162

#SPJ1

6 0
2 years ago
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