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3241004551 [841]
2 years ago
10

Easy problem for you guys to solve

Mathematics
1 answer:
Zepler [3.9K]2 years ago
6 0

In the ΔIJH, the value of the cosec (I) is 1\frac{9}{56}.

Given ΔIJH the length of the hypotenuse is 65,  the length of the base is 33, and the length of the opposite side is 56.

We have to find the value of the cosec (I).

A function of an arc or angle that is most easily represented in terms of the ratios of pairs of sides of a right-angled triangle, such as the sine, cosine, tangent, cotangent, secant, or cosecant.

We know cosec (I) = hypotenuse / opposite side

Substitute the values

cosec (I) = 65/56

              = 1\frac{9}{56}

Hence the value of cosec (I) is 1\frac{9}{56}.

Learn more about  trigonometric function here: brainly.com/question/24349828

#SPJ10

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Can you answer this one for my plz I need before Monday ty
ira [324]
Since you need the vertex, and knowing the vertex will tell you the axis of symmetry, it is convenient to put the equation in vertex form.
.. -x^2 -4x +1
.. = -(x^2 +4x) +1
.. = -(x^2 +4x +(4/2)^2) +1 -(-(4/2)^2) . . . . complete the square
.. = -(x +2)^2 +5

The vertex is (-2, 5).
The axis of symmetry is x = -2.

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3 years ago
Pls help 50 points and brainliest
daser333 [38]

Answer:

variable equation = 9h = 72

regular equation= 72/9=8

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Step-by-step explanation:

8 0
2 years ago
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Simplify the difference quotient f(x) − f(a) / x − a if x ≠ a. f(x) = x^3 − 36
Annette [7]
Fx - fa÷x - ( aif x ÷ n (2.178282)) (a) = -2.718282a² fix²+ fnx² - afn ÷ nx
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What would be the missing angle?
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Simplify f+g / f-g when f(x)= x-4 / x+9 and g(x)= x-9 / x+4
steposvetlana [31]

f(x)=\dfrac{x-4}{x+9};\ g(x)=\dfrac{x-9}{x+4}\\\\f(x)+g(x)=\dfrac{x-4}{x+9}+\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)+(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2+x^2-9^2}{(x+9)(x+4)}=\dfrac{2x^2-16-81}{(x+9)(x+4)}=\dfrac{2x^2-97}{(x+9)(x+4)}\\\\f(x)-g(x)=\dfrac{x-4}{x+9}-\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)-(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2-(x^2-9^2)}{(x+9)(x+4)}=\dfrac{x^2-16-x^2+81}{(x+9)(x+4)}=\dfrac{65}{(x+9)(x+4)}


\dfrac{f+g}{f-g}=(f+g):(f-g)=\dfrac{2x^2-97}{(x+9)(x+4)}:\dfrac{65}{(x+9)(x+4)}\\\\=\dfrac{2x^2-97}{(x+9)(x+4)}\cdot\dfrac{(x+9)(x+4)}{65}\\\\Answer:\ \boxed{\dfrac{f+g}{f-g}=\dfrac{2x^2-97}{65}}

6 0
3 years ago
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