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4 years ago

Solve for q - 3/4q = 12 q = ___

2 answers:

3 0

$\text{We know: } \\ \boxed{\frac{\ -3\ }{\ 4\ }*q=12 } \\ \\ q=12: \frac{\ -3\ }{\ 4\ } \\ \\ q=12* \frac{\ 4\ }{\ -3\ } \\ \\ q= \frac{\ 48\ } {\ -3\ } \\ \\ q=-16 \\ ********** \\ \\ \text{Check:}\\ \\ \boxed{ \frac{-3}{4}*-16=12 } \\ \\ \frac{48}{4} =12 \ TRUE$

vfiekz [6]4 years ago

3 0

$\frac{-3}{4q}=12$
Multiply by q to eliminate var in denominator
$\frac{ \not q*(3)}{4 \not q} =q*12$
$q= \frac{-3}{4} =q*12$
Common denominator is 4
$\frac{ \not 4(-3)}{ \not 4} =4q*12$
$-3=4q*12$
$-3=48q$
Divide by 48
$q=\frac{ \not 3}{ \not 4 \not 8}$
$q=-16$
The answer is: $q=-16$

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We can draw a unit circle as:

The sine is the vertical leg (or vertical projection) of the triangle formed by the angle.

The cosine will be the horizontal leg (or horizontal projection) of this same triangle.

The hypotenuse of this triangle will be the radius if the unit circle (r = 1).

a) The sine will be increasing from θ = 0° until it reaches its maximum at 90°.

Then it will be decreasing for angles between 90° and 270°.

Between 270° and 0° it will be increasing again.

b) The cosine has a maximum value for an angle of 0°.

It will decrease until 180°.

Then it will be increasing from angles between 180° and 360°.

c) We can see in the unit circle that the sine is 0 when the angle is one the horizontal axis, like when θ = 0° or θ = 180°.

The cosine will be equal to 0 when the angle is vertical: θ = 90° and θ = 270°.

Answer:

a) Increasing: (0°, 90°) and (270°, 360°)

Decreasing: (90°, 270°)

b) Increasing: (180°, 360°)

Decreasing: (0°,180°).

c) Sin(θ) = 0 for θ = 0° and θ = 180°

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2 years ago

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Alchen [17]

Answer:

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The area of a rectangular ceiling tile is 756 square inches. The perimeter is 114 inches. What are the dimensions of the tile?

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Here we need to use what we know about rectangles to make a system of equations.

By solving that system we found that the tile has a length of 36 inches and a width of 21 inches.

Remember that for a rectangle of length L and width W, the perimeter is:

P = 2*(L + W)

And the area is:

A = W*L

Here we know that the perimeter is 114 inches, then we can write:

114in = 2*(L + W)

We also know that the area is 756 in^2, then we can write:

756 in^2 = L*W

So we found two equations, which means that we have a system of two equations with two variables:

114in = 2*(L + W)

756 in^2 = L*W

To solve this, the first step is to isolate one of the variables in one of the equations, we can isolate L in the first equation:

114in = 2*(L + W)

114in/2 = (L + W)

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57in - W = L

Now that we have an expression equivalent to L, we can replace it in the other equation to get:

756 in^2 = L*W

756 in^2 = (57in - W)*W

Now we can solve this for W.

756 in^2 = W*57in - W^2

W^2 - W*57in + 756 in^2 = 0

The solutions are given by the Bhaskara's formula:

$W = \frac{57in \pm \sqrt{(-57in)^2 - 4*1*(756in^2)} }{2*1} = \frac{57in \pm 15in}{2}$

Then the two possible values of the width will be:

W = (57in + 15in)/2 = 36 in

W = (57in - 15in)/2 = 21 in

Suppose that we choose the second solution, W = 21in

Now using the equation 57in - W = L we can find the value of L

L = 57in - W = 57in - 21in = 36in

L = 36in

Then we found that the tile has a length of 36 inches and a width of 21 inches.

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