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Paha777 [63]
2 years ago
10

Please help, 30 points, will give brainly

Mathematics
1 answer:
zhenek [66]2 years ago
3 0

Answer:

<u>∠ABC = 39°</u>

Step-by-step explanation:

Since ED bisects ∠CBD :

<u>∠EBD = ∠CBE = 30°</u>

<u />

Now, <u>∠ABD = ∠ABC + ∠CBE + ∠EBD = 99°</u>

Solving :

  • 99° = ∠ABC + 30° + 30°
  • ∠ABC = 99° - 60°
  • <u>∠ABC = 39°</u>
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Which Greek mathematician wrote the most definitive text on geometry, one that is still referred to today?
frutty [35]
The Greek mathematician who wrote the most definitive text on geometry, one that is still referred to today is Euclid. Euclid of Alexandria is also called "Father of Geometry". He was a great Greek mathematician. Euclidean geometry is still widely taught in schools and colleges. I hope the answer comes to your help.
5 0
2 years ago
Which function is the same as y = 3 cosine (2 (x startfraction pi over 2 endfraction)) minus 2? y = 3 sine (2 (x startfraction p
kirza4 [7]

The function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

<h3>How to convert sine of an angle to some angle of cosine?</h3>

We can use the fact that:

\sin(\theta) = \cos(\pi/2 - \theta)\\\sin(\theta + \pi/2) = -\cos(\theta)\\\cos(\theta + \pi/2) = \sin(\theta)

to convert the sine to cosine.

<h3>Which trigonometric functions are positive in which quadrant?</h3>
  • In first quadrant (0 < θ < π/2), all six trigonometric functions are positive.
  • In second quadrant(π/2 < θ < π), only sin and cosec are positive.
  • In the third quadrant (π < θ < 3π/2), only tangent and cotangent are positive.
  • In fourth (3π/2 < θ < 2π = 0), only cos and sec are positive.

(this all positive negative refers to the fact that if you use given angle as input to these functions, then what sign will these functions will evaluate based on in which quadrant does the given angle lies.)

Here, the given function is:

y= 3\cos(2(x + \pi/2)) - 2

The options are:

  1. y= 3\sin(2(x + \pi/4)) - 2
  2. y= -3\sin(2(x + \pi/4)) - 2
  3. y= 3\cos(2(x + \pi/4)) - 2
  4. y= -3\cos(2(x + \pi/2)) - 2

Checking all the options one by one:

  • Option 1: y= 3\sin(2(x + \pi/4)) - 2

y= 3\sin(2(x + \pi/4)) - 2\\y= 3\sin (2x + \pi/2) -2\\y = -3\cos(2x) -2\\y = 3\cos(2x + \pi) -2\\y = 3\cos(2(x+ \pi/2)) -2

(the last second step was the use of the fact that cos flips its sign after pi radian increment in its input)
Thus, this option is same as the given function.

  • Option 2: y= -3\sin(2(x + \pi/4)) - 2

This option if would be true, then from option 1 and this option, we'd get:
-3\sin(2(x + \pi/4)) - 2= -3\sin(2(x + \pi/4)) - 2\\2(3\sin(2(x + \pi/4))) = 0\\\sin(2(x + \pi/4) = 0

which isn't true for all values of x.

Thus, this option is not same as the given function.

  • Option 3: y= 3\cos(2(x + \pi/4)) - 2

The given function is y= 3\cos(2(x + \pi/2)) - 2 = 3\cos(2x + \pi) -2 = -3\cos(2x) -2

This option's function simplifies as:

y= 3\cos(2(x + \pi/4)) - 2 = 3\cos(2x + \pi/2) -2 = -3\sin(2x) - 2

Thus, this option isn't true since \sin(2x) \neq \cos(2x) always (they are equal for some values of x but not for all).

  • Option 4: y= -3\cos(2(x + \pi/2)) - 2

The given function simplifies to:y= 3\cos(2(x + \pi/2)) - 2 = 3\cos(2x + \pi) -2 = -3\cos(2x) -2

The given option simplifies to:

y= -3\cos(2(x + \pi/2)) - 2 = -3\cos(2x + \pi ) -2\\y = 3\cos(2x) -2

Thus, this function is not same as the given function.

Thus, the function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

Learn more about sine to cosine conversion here:

brainly.com/question/1421592

4 0
2 years ago
Read 2 more answers
Which equation is true when n = 4?<br> A<br> 2n = 6<br> Bn+3 = 7<br> 9- n = 13
kari74 [83]

Answer:

The answer is B. n+3=7

Step-by-step explanation:

^^^^^^^

5 0
2 years ago
Read 2 more answers
What 5.00 times 0.75 equal
xeze [42]
We know that 75=5*5*3 so
5*5*3 times 5=125*3=375 then move the decimal place two places to the left (divide by 100) and get 3.75

or you could note that 0.75=3/4 so
3/4 times 5=
15/4
simplified=3 and 3/4
3 0
3 years ago
I need to use the Pythagorean Theorem to find X. I don’t know how to do that!!
frutty [35]

Answer:

17) x=8

18) x=\sqrt{189} or x= 3\sqrt{21}

Step-by-step explanation:

So the rule is a^{2} +b^{2} =c^{2}, "c" being the hypothenuse, or the long line that is opposite to the right angle.

17) We know that both values of x are equal to each other, which makes everything 10x easier!

                                               x^{2} +x^{2} =(8\sqrt{2})  ^{2}

(by the way we know the x values are our a and b values because they are legs! the way I like to remember the legs is that they are connected to the right angle box, and therefore support the hypothenuse)

<em>simplify</em> (╥︣﹏╥)    

                                                2x^2=8^2(\sqrt{2} )^{2}  

                                                  2x^2=64(2)

                                                   2x^2=128    

                                                  x^{2}= 128/2  

                                                     x^{2} =64

                                                        x=8      

18) Just pretend that the flipped triangle doesn't exist. It's parallel to the other triangle with values on it, and basically servers no purpose other than being parallel to the sister triangle :)

Anyways, since we know the hypothenuse (15) but we don't know one of our leg values (x), we're going to change our equation a bit!

                                                    c^{2} - b^{2} = a^{2}

It doesn't matter if you put the one leg value in a or b, just as long as you stick to that same equation you started with the entire time!

                                                   15^{2} -6^{2} =x^{2}

                                                   225-36=x^{2}  

                                                     x^{2} =189

                                               x=\sqrt{189}=3\sqrt{21}

        The more you do these, the easier they'll get, so don't worry!

                           

4 0
3 years ago
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