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3 years ago

Pls help I’m timed thanks

Mathematics

1 answer:

alexgriva [62]3 years ago

3 0

Answer:

Picture 1: b

Picture 2: d

Picture 3: c

Step-by-step explanation:

Hi there,

For Picture 1, to solve for an inverse function, just simply solve for x. Since the closest you can get is arccos(πx)=2y+6, you just

solve for cosine as a function: cos(2y+6)=πx and solving for x:

x= (1/π)cos(2y+6) but since variable choice is arbitrary, you can now redefine y and x:

y= (1/π)cos(2x+6)

For Picture 2, tanx is equivalent to sinx/cosx and cscx is just the reciprocal of sinx. So, it becomes:

$\frac{sinx}{cosx} *\frac{1}{sinx} =\frac{1}{cosx} =cscx$

We have already been giving the cosine value of 2, and its inverse is thus 1/2.

For Picture 3, I would recommend revisiting polar coordinates.

Polar coordinates are in the form (r, θ).

$r = \sqrt{x^{2} +y^{2} } }$ and θ $=arctan(\frac{y}{x})$ . Recognize there are two possible radii, depending on what side of the circle you start from!

thanks,

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Which measurement is the width of this rectangle? Area = 112 cm²

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C. 10 cm
To understand the given problem use the formular of area.

Hence, area is the length and width of the object in unit squared.

Mathematically expressing, a = l x w
Since a = l x w
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3 years ago

The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base and PA = 2sqrt5 in. The base edges AB = 10 in, AC

DIA [1.3K]

Answer:

PQ = √84 = 2√21 in ≈ 9.165 in

Step-by-step explanation:

The base edges AB = 10 in, AC = 17 in, and BC = 21 in

Q ∈ BC and AQ is the altitude of the base.

Let BQ = x in ⇒ CQ = (21 - x) in

ΔAQB a right triangle at Q

∴ AQ² = AB² - BQ² = 10² - x² ⇒(1)

ΔAQC a right triangle at Q

∴ AQ² = AC² - CQ² = 17² - (21-x)² ⇒(2)

Equating (1) and (2)

∴ 10² - x² = 17² - (21-x)²

∴ 10² - x² = 17² - (21² - 42x + x²)

∴ 10² - x² = 17² - 21² + 42x - x²

∴ 10² - 17² + 21² = 42x

∴ 42x = 252

∴ x = 252/42 = 6

Substitute at (1)

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∵ The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base

∴ PA⊥AQ

∴ ΔPAQ is a right triangle at A,

PA = 2sqrt5 in and AQ = 8 in

∴ PQ (hypotenuse) = √(PA² + AQ²)

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3 years ago

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Answer:

Possibility 1: $\frac{(x+1)(x+1)}{(x-4)(x+5)}$

Possibility 2: $\frac{(x+1)(x+3)(x-3)}{x(x-4)(x+5)}$

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There are infinitely many more possibilities.

Step-by-step explanation:

So we are looking for a fraction in terms of x.

We have a horizontal asymptote so that means the degree of the top has to equal to degree of the bottom. It is also at y=1 which means the coefficient of the leading term on top and bottom must be the same (but not zero) since the same number divided by the same number is 1.

Now we also have a vertical asymptote at x=4 which means we need a factor of x-4 on bottom.

So there is a lot of possibilities. Here are a few:

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You have the degrees are the same on top and bottom and the leading coefficients are the same. You also have that factor of (x-4) on bottom.

Possibility 2: $\frac{(x+1)(x+3)(x-3)}{x(x-4)(x+5)}$

Possibility 3: $\frac{-4(x-4)(x+1)}{-4(x-4)(x-4)}$

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Answer:

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