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Ostrovityanka [42]
3 years ago
13

I just need help solving this problem. Also please don't put the answer into a file because I can't open them. It's also really

sketchy.

Mathematics
1 answer:
dem82 [27]3 years ago
5 0

Answer:

<em>31° </em>

Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=tanx%5E%7B2%7D%20-%5Cfrac%7B1%7D%7B3%7D%20%3D0" id="TexFormula1" title="tanx^{2} -\frac{1}{3}
lianna [129]

These are expression written as a function of sine, cosine and tangent. The value of x from the given function is 4.29

Trigonometry expression

These are expression written as a function of sine, cosine and tangent.

Given

tanx² - 1/3 = 0

Add 1/3 to both sides

tanx² - 1/3 + 1/3 = 0 + 1/3

tanx² = 1/3

x² = arctan(1/3)

x² = 18.44

x =4.29

Hence the value of x from the given function is 4.29

Learn more on trig function here: brainly.com/question/24349828

#SPJ2

6 0
2 years ago
Read 2 more answers
Find the missing leg of a triangle with legs of 7 and a hypotenuse of 10<br> Help please
ahrayia [7]

Answer:

i believe its 24 in for the other leg

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Which phrase represents this expression?
quester [9]

Answer:

the correct answer is 3 times the difference of 62 and 42

7 0
3 years ago
If x &gt; 2, find lx-2l<br><br> Help pleaseee
Lyrx [107]
<h3>Answer:  x-2</h3>

Explanation:

If x > 2, then x-2 > 0 after subtracting 2 from both sides.

Since x-2 is always positive when x > 2, this means the absolute value bars around the x-2 aren't needed. The results of |x-2| and x-2 are perfectly identical.

For example, if we tried something like x = 5, then

  • x-2 = 5-2 = 3
  • |x-2| = |5-2| = |3| = 3

Both outcomes are 3. I'll let you try other x inputs.

So because |x-2| and x-2 are identical, this means |x-2| = x-2 for all x > 2.

In short, we just erase the absolute value bars.

8 0
3 years ago
PLEASE PLEASE HELP<br><br>question is attached
Andreas93 [3]
Short Answer
x1 = 4.8284
x2 = - 0.828
Remark
Substitute the value for y from the first equation into the second equation. Multiply by 4 and then see if it factors out. Solve for x first and then y. 

Step one 
Solve for y in the first equation. Subtract x from both sides.
y = 2 - x

Step Two
Equate the two ys.
2 - x = - 1/4x^2 + 3

Step Three
Bring the left side over to the right side.
0 = -1/4 x^2 + x + 3 -  2 Combine the like terms.
0 = -1/4 x^2 + x + 1      

Step Four
0 = -1/4 x^2 + x + 1      Multiply through by 4
0 = - x^2 + 4x + 4

Step five
This won't factor. The only thing you can do is use the quadratic equation for roots.

a = - 1
b = 4
c = 4

\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac} }{2a}

\text{x = }\dfrac{ -(4) \pm \sqrt{4^{2} - 4(-1)4}}{-2}

\text{x = }\dfrac{ -(4) \pm \sqrt{16 + 16}}{-2}

\text{x = } 2 \pm \sqrt{32 }\div -2

x = 2 +/- sqrt(4^2 * 2) / (- 2)

x = 2 +/- 4 sqrt(2) / - 2
x = 2 -/+ 2 sqrt(2)
x = 2 -/+ 2 *(1.414) 
x = 2 -/+ 2.828

x1 = 4.8284
x2 = - 0.828
4 0
4 years ago
Read 2 more answers
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