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Yuri [45]
3 years ago
12

Write the expression in algebraic form. [hint: sketch a right triangle.] tan(arcsec(x/3))

Mathematics
1 answer:
WARRIOR [948]3 years ago
6 0

Sketch  a right triangle having adjacent side(A) is given as “3”, hypotenuse side (H) is “x” and assigning angle “a” as the angle between A and H. Using Pythagorean theorem, you will get “square root of x-squared minus 9” as the opposite side (O). Using SOH CAH TOA function, and since secant is the reciprocal of cosine, sec(a) = x/3. Thus, a = arcsec(x/3). The remaining expression tan(a) is Opposite side over Adjacent side which is equal to “square root of x^2 - 9” over "3". Therefore, the algebraic expression would be: tan(arcsec(x/3)) = “sqrt (x^2 -9)” /3. Different answers can be made depending on which side you consider the “3” and “x”.

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3 years ago
What function is represented by a line with slope -2 that passes through the point (0,4)
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We can write the function in slope-intercept form (y + mx + b). Since we have the slope, we can solve for the y-intercept.

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Input the data we have now.

y = -2x + 4

Since we are dealing with a function switch out the y for f(x)

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3 years ago
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Alexeev081 [22]

Answer:

True

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3 years ago
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4 0
3 years ago
Read 2 more answers
Solve the equation 25x²+100x+200=0.
k0ka [10]

Answer:

x = -2+2i ,x = -2-2i

Step-by-step explanation:

given,

equation 25x²+100x+200=0

dividing the equation by 25 on both side

x²+4x+8=0

using

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

x = \dfrac{-4\pm \sqrt{4^2-4\times 1\times 8}}{2}

x = \dfrac{-4+ \sqrt{4^2-4\times 1\times 8}}{2},x = \dfrac{-4-\sqrt{4^2-4\times 1\times 8}}{2}

x = \dfrac{-4+ \sqrt{-16}}{2},x = \dfrac{-4-\sqrt{-16}}{2}

x = \dfrac{-4+ 4i}{2},x = \dfrac{-4-4i}{2}

x = -2+2i ,x = -2-2i

6 0
3 years ago
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