Which function has a vertex at ( 2 , -5 )?

Sec B cot B = sec B csc B cot B tan B

irinina [24]

Given:
tan(B/2) = sec(B) / (sec(B) * csc(B) + csc(B)) 

Apply the half angle formula to convert tan(B/2) to terms of B: 
sin(B) / (1+cos(B)) = sec(B) / (sec(B) * csc(B) + csc(B)) 

Convert everything else to be in terms of sin and cos: 
sin(B) / (1+cos(B) = (1/cos(B)) / ((1/cos(B)) * (1/sin(B)) + (1/sin(B))) 

Multiply right side by "sin(B)/sin(B)" to simplify the fractions: 
sin(B) / (1+cos(B) = (sin(B)/cos(B)) / ((1/cos(B)) + 1) 

Change "1" to cos(B)/cos(B) and then combine over 
common denominator: 
sin(B) / (1+cos(B) = (sin(B)/cos(B)) / ((1/cos(B)) + cos(B)/cos(B)) 
sin(B) / (1+cos(B) = (sin(B)/cos(B)) / ((1+cos(B))/cos(B)) 

Dividing by a fraction equals multiplying by its reciprocal: 
sin(B) / (1+cos(B) = (sin(B)/cos(B)) * (cos(B) / (1+cos(B))) 

Multiply terms on the right side (canceling cos(B) terms): 
sin(B) / (1+cos(B) = sin(B) / (1+cos(B))

6 0

4 years ago

Read 2 more answers

A rectangular sheet of cooper is twice as long as it is wide. From each corner a 3-inch square is cut out, and the ends are then

Galina-37 [17]

Answer:

length=24 inches and width=12 inches

Step-by-step explanation:

Let width of the rectangular sheet be x.

Then its length is twice as width = 2x

From each corner 3-inch square is cut to form a tray.

The length of the tray formed = l = 2x-6 (total length - length of 2 squares cut out from both ends)

The width of the tray formed = b = x-6 (total width - length of 2 squares cut out from both ends)

Height of the tray formed = h = 3 inches

Volume of the tray = l*b*h = (2x-6)*(x-6)*3 = 324 cubic inches.

$2x^{2}-18x+36=108\\x^{2}-9x-36=0\\(x-12)(x+3)=0\\x=12$

The length and width of the sheet is 24 and 12 inches respectively.

6 0

3 years ago

Which of the following represents the graph of f(x) = 2x − 3? graph begins in the second quadrant near the line equals negative

Snezhnost [94]

Answer:

The second choice: graph begins in the second quadrant near the axis and increases slowly while crossing the ordered pair 3, 1. The graph then begins to increase quickly throughout the first quadrant.

Explanation:

1) As per the set of choices, the function is:

$f(x)=2^{x-3}$

2) Therefore it is an exponential function with these characteristics:

Since, the bases is greater than 1 (2), the function is growing in all the domain.
The domain is all real numbers (- ∞, ∞).
To know where the function starts, calculate the limit of f(x) as x trends to negative infinity:

$\lim_{x \to \infty} 2^{x-3}=0^{+}$

That means that the range is y > 0, and so the graph starts in the second quadrant.

You can find the y-intersection making x = 0, which is 2⁰ ⁻ ³ = 2 ⁻³ = 1/8 = 0.25. So, the graph cross the y axis at y = 0.25.
That tells you, that the function increases slowly, at least, until that point (0, 0.25).
The other bullet point is when x = 3: 2 ³ ⁻ ³ = 2⁰ = 1. Therefore, the graph passes through the point (3, 1).
From that point, the function starts to increase rapidly (since it is exponential).
Those are the characteristics given by the second choice: graph begins in the second quadrant near the x-axis and increases slowly while crossing the ordered pair 3, 1. The graph then begins to increase quickly throughout the first quadrant.

You surely will find useful to watch the graph that I have attached.

4 0

3 years ago

Distance between the points

Harman [31]

To understand the distance formula, you first need to understand the Pythagorean Theorem. For a refresher, the theorem states that the square of the legs of a right triangle is equal to the the square of its hypotenuse (the side opposite the right angle), or in symbols:

$a^2+b^2=c^2$ , where a and b are the lengths of the legs, and c is the length of the hypotenuse. In the context of the x-y plane, the legs of the triangle correspond to separate x and y values on the plane, and the hypotenuse corresponds to a straight line between two points on that plane.

To find the distance between the points you've listed, (2√5,4) and (1,2√3), we'll first need to find the "legs" of the triangle. To find the length of the x leg, we'll just need the distance between the x values of the points, which we find to be 2√5-1. We do the same for the y component, which ends up being 4-2√3. Now that we have our legs, we're ready to find the hypotenuse - or the distance.

Going back to Pythagorus's equation, we have:

$(2 \sqrt{5}-1)^{2}+(4-2 \sqrt{3})^{2}=d^2$

where d, the hypotenuse of the triangle, means "distance."

To solve for d, we take the square root of both sides:

$d= \sqrt{(2 \sqrt{5}-1)^2+(4-2 \sqrt{3} )^2}$

And from there, all that's left to do is solve the right side of the equation, which just ends up being rote calculation.

Edit: I'll go through the steps of that calculation here. We'll start by expanding each of the squared terms inside the radical:

$(2 \sqrt{5}-1)^2=(2 \sqrt{5}-1)(2 \sqrt{5}-1)=(2 \sqrt{5}-1)2 \sqrt{5}-(2 \sqrt{5}-1)$
$=(2 \sqrt{5})^2-2 \sqrt{5}-2 \sqrt{5}+1=20-4\sqrt{5}+1$

$(4-2\sqrt{3})^2=(4-2\sqrt{3})(4-2\sqrt{3})=(4-2\sqrt{3})4-(4-2\sqrt{3})2\sqrt{3}$
$=16-8\sqrt{3}-8\sqrt{3}+(2\sqrt{3})^2=16-16\sqrt{3}+12$

Putting those values back under the radical:

$\sqrt{20-4\sqrt{5}+1+16-16\sqrt{3}+12}$

Collecting constants:

$\sqrt{49-4\sqrt{5}-16\sqrt{3}}$

If you wanted an exact answer, this messy-looking thing would be it, and you can verify those results on WolframAlpha if you'd like. If you want an approximation, just enter that expression in to the online calculator of your choice, and it should give out the value of approx. 3.51325.

In general, if you want to solve for the distance between two points $(y_{1},x_{1})$ and $(y_{2},x_{2})$ , the formula is:

$d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

4 0

3 years ago

2. If the measure of angle 1 is 38 degrees, what is the measure of angle 2?

noname [10]

The answer is C Thank

6 0

3 years ago