There are 10 students on the track team who compete in sprinting events. They make up 25% of the track team. How many students are on the track team?

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Answer:

  44.66 in

Step-by-step explanation:

The side opposite the marked angle is given, and the side adjacent to it is the one wanted. The relevant trig relation is ...

  Tan = Opposite/Adjacent

Solving for the Adjacent side, we find ...

  Adjacent = Opposite/Tan

  PQ = (29 in)/tan(33°) ≈ 44.66 in

Answer:

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Answer:

f(g(x)) = 9x^2 + 15x - 6

Step-by-step explanation:

We are using function g(x) = 3x - 1 as the input to function f(x) = x^2 + 7x.

Starting with f(x) = x^2 + 7x, substitute g(x) for x on the left side and likewise substitute x^2 + 7x for each x on the right side.  We obtain:

f(g(x)) = (3x - 1)^2 + 7(3x - 1).

If we multiply this out, we get:

f(g(x)) = 9x^2 - 6x + 1 + 21x - 7, or

f(g(x)) = 9x^2 + 15x - 6

Answer:

W is multiplied by 8

Step-by-step explanation:

If W varies jointly with x, y, and z, we can say that

W = k (xyz), with k being a constant for our original equation. We are asked what will happen to W if x, y, and z are each doubled. To figure this out, we can go back to our equation,

W = k (xyz)

First, we can double x, meaning that we multiply it by 2. Doing this, we get

W = k (2x * y * z)

Then, we can double y and z in a similar fashion, resulting in

W = k (2x * 2y * 2z)

W = 8 * k (xyz)

The new W, after all the doubling, is equal to 8 * k * x * y * z. The old W is equal to k * x * y * z. It can be determined that the new W is equal to 8 * the old W, so W is multiplied by 8

Answer:

50% growth would would be (1 + .5)^ n

and 100% growth would be (1+ 1)^n

I assume that the answer would be (1+3)^n

for 300% growth the factor would be 3

Step-by-step explanation:

Answer:

a

Step-by-step explanation:

Answer:

V ≈ 615.75

r Radius 7

h Height 4

Answer:

Casey's monthly charge for making 1,100 minutes of calls is $70.

Step-by-step explanation:

We can write a piecewise function to model the situation.

Since Casey's phone service only charges a monthly fee of $30 for the first 1000 minutes, we can write that for calling t minutes:

[tex]\displaystyle C(t) = 30\text{ if } t\leq 1000[/tex]

In other words, the total cost is only $30 is the total minutes of call is less than  1000 minutes.

However, if the total minutes of calls is greater than 1000, then its $0.40 per minute on top of the 30. Thus:

[tex]\displaystyle C(t) = 30 + 0.4(t-1000)\text{ if } t>1000[/tex]

All together, our piecewise function will be:

[tex]\displaystyle C(t) = \begin{cases} 30 & t\leq 1000 \\ 30 + 0.4(t-1000) & t>1000\end{cases}[/tex]

We want to determine Caseys monthly charge if he makes 1,100 minutes of calls. So, t = 1100. Since 1100 > 1000, we will use the second equation. This yields:

[tex]C(1100)= 30+0.4((1100)-1000)[/tex]

Evaluate:

[tex]\displaystyle C(1100) = 30+0.4(100) = 30+40=\$70[/tex]

Casey's monthly charge for using 1,100 minutes of call is $70.

Answer:

Step-by-step explanation:

Answer:

e. 1.6π cm²

Step-by-step explanation:

Since the window is in a semi-circular shape, its area A = πD²/4 ÷ 2 = πD²/8 where D = diameter of window = 64 cm

Now, the error in the area dA = dA/dD × dD where dD = error in the diameter = 0.1 cm and dA/dD = derivative of A with respect to D.

So, dA/dD = d(πD²/8)/dD = 2 × πD/8 = πD/4

So, the differential dA = dA/dD × dD

dA =  πD/4 × dD

Substituting D = 64 cm and dD = 0.1 cm into the equation, we have

dA =  πD/4 × dD

dA =  π × 64 cm/4 × 0.1 cm

dA =  π × 16 cm × 0.1 cm

dA =  π × 1.6 cm²

dA = 1.6π cm²

So, the maximum error in computing the area of the window is 1.6π cm²

9514 1404 393

Answer:

  y = -x +3

  m = -1

  b = 3

Step-by-step explanation:

To solve the given equation for y, divide by the coefficient of y.

  (-3y)/(-3) = (3x -9)/(-3)

  y = -x +3

__

The slope is the x-coefficient, M = -1.

__

The y-intercept is the added constant, B = 3.

__

Both equations are graphed in the attachment. Texture has been added to the original so you can see the graphs are the same line.

3 1/5 = y - 12/25

31/5+y = -12/25

31+y = -12/25×5

31+y = -12/5

y = -12/5-31

y = -143/5

y = -28.6

HOPE IT HELPS

Answer:

Step-by-step explanation:

50%

Answer:

50%

Step-by-step explanation:

Win % = wins / total * 100%

Win% = (5/(5 + 5)) * 100

Win% = 5/10 * 100 = 50%

Answer:

70.908 is greater than 7.908 .

I hope your day goes nice

Answer:

both are different because of different placement of point.

Answer:

D

Step-by-step explanation:

0.0015

hope this helps

Answer:

b = 53

c = 53

Step-by-step explanation:

Alright so we already know that b and c are going to have the same angle measures. We can find b by subtracting 180 degrees to 127. Why you may ask? Its because when b and 127 are added together its obvious that it creates a straight line (supplementary angle). This means that two angles will sum up to 180 degrees. We can create an easy equation and solve for b.

127 + b = 180

b = 53 and c = 53

Best of Luck!

Step-by-step explanation:

Let find the distance of the diameter. using distance formula.

[tex](3 + 1) {}^{2} + ( - 2 + 6) {}^{2} = \sqrt{8} [/tex]

The diameter is sqr root of 8 units.

A circle equation is

[tex] {x}^{2} + {y}^{2} = {r}^{2} [/tex]

where r is the radius. The radius is half the diameter so

[tex]r = \frac{ \sqrt{8} }{2} = \frac{ \sqrt{8} }{ \sqrt{4} } = \sqrt{2} [/tex]

[tex] {r}^{2} = { \sqrt{2} }^{2} = 2[/tex]

So our radius is 2.

Now we need to find the midpoint or Center of the diameter.

[tex] \frac{ - 6 - 2}{2} = - 4[/tex]

[tex] \frac{3 - 1}{2} = 1[/tex]

So the center of the circle is (-4,1). So our equation of the Circle us

[tex](x + 4) {}^{2} + (y - 1) {}^{2} = ( \sqrt{2} ) {}^{2} [/tex]

9514 1404 393

Answer:

  (d)  27.4%

Step-by-step explanation:

The desired percentage is ...

  (juniors for Kato)/(total juniors) × 100%

  =  129/(129 +194 +147) × 100%

  = (129/470) × 100% ≈ 27.4%

About 27.4% of juniors voted for Kato.

Answer:

B

Step-by-step explanation:

Pull in -4 and 0 into the variable x. and the 4 and -2 in the variable y and solve to get an equal answer.

Answer:

was assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?

1. In the game of baseball, which is rather boring, the batter stands on home base (Hughes 1).

2. Baseball is rather surprisingly known as "America's Favorite Pastime."

3. Baseball is "a rather boring sport played between two teams of nine players" (Hughes 1).

4. All of these are plagiarism.

The answer tells that only the third choice is NOT a plagiarism. My question is, why is the first choice a plagiarismwas assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?

1. In the game of baseball, which is rather boring, the batter stands on home base (Hughes 1).

2. Baseball is rather surprisingly known as "America's Favorite Pastime."

3. Baseball is "a rather boring sport played between two teams of nine players" (Hughes 1).

4. All of these are plagiarism.

The answer tells that only the third choice is NOT a plagiarism. My question is, why is the first choice a plagiarismwas assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?

1. In the game of baseball, which is rather boring, the batter stands on home base (Hughes 1).

2. Baseball is rather surprisingly known as "America's Favorite Pastime."

3. Baseball is "a rather boring sport played between two teams of nine players" (Hughes 1).

4. All of these are plagiarism.

The answer tells that only the third choice is NOT a plagiarism. My question is, why is the first choice a plagiarismwas assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?

1. In the game of baseball, which is rather boring, the batter stands on home base (Hughes 1).

2. Baseball is rather surprisingly known as "America's Favorite Pastime."

3. Baseball is "a rather boring sport played between two teams of nine players" (Hughes 1).

4. All of these are plagiarism.

The answer tells that only the third choice is NOT a plagiarism. My question is, why is the first choice a plagiarismwas assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?

1. In the game of baseball, which is rather boring, the batter stands on home base (Hughes 1).

2. Baseball is rather surprisingly known as "America's Favorite Pastime."

3. Baseball is "a rather boring sport played between two teams of nine players" (Hughes 1).

4. All of these are plagiarism.

The answer tells that only the third choice is NOT a plagiarism. My question is, why is the first choice a plagiarismwas assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?

1. In the game of baseball, which is rather boring, the batter stands on home base (Hughes 1).

2. Baseball is rather surprisingly known as "America's Favorite Pastime."

3. Baseball is "a rather boring sport played between two teams of nine players" (Hughes 1).

4. All of these are plagiarism.

The answer tells that only the third choice is NOT a plagiarism. My question is, why is the first choice a plagiarismwas assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?

Answer:

[tex]15y^9 - 7y^3 + y[/tex]

Nonic polynomial

Step-by-step explanation:

Given

[tex]y - 7y^3 + 15y^9[/tex]

Required

Write in standard form

The standard form of a polynomial is:

[tex]ay^n + by^{n-1} + ......... + k[/tex]

So, we have:

[tex]y - 7y^3 + 15y^9[/tex]

The standard form is:

[tex]15y^9 - 7y^3 + y[/tex]

And the name is: Nonic polynomial (because it has a degree of 9)

9514 1404 393

Answer:

Step-by-step explanation:

The value of d is the average of the maximum and the minimum.

  d = (-4 +-5)/2 = -9/2 = -4.5

The value of 'a' is the difference between the maximum and d.

  a = -4 -(-4.5) = 0.5

The value of b is 2π divided by the period. Here, the half-period is 3.5 -(-1) = 4.5, so ...

  b = 2π/9 = π/4.5

The value of c is the value that makes the cosine argument zero at x=3.5.

  (π/4.5)(3.5) +c = 0

  c = -7π/9

Answer:

Placements are basically extended internships or work experience assignments

Step-by-step explanation:

Answer:  [tex]-2+\sqrt{3}[/tex]

=========================================================

Work Shown:

Apply the following trig identity

[tex]\tan(A - B) = \frac{\tan(A)-\tan(B)}{1+\tan(A)*\tan(B)}\\\\\tan(225 - 60) = \frac{\tan(225)-\tan(60)}{1+\tan(225)*\tan(60)}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+1*\sqrt{3}}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\[/tex]

Now let's rationalize the denominator

[tex]\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\\tan(165) = \frac{(1-\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}\\\\\tan(165) = \frac{(1-\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{(1)^2-2*1*\sqrt{3}+(\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{1-2\sqrt{3}+3}{1-3}\\\\\tan(165) = \frac{4-2\sqrt{3}}{-2}\\\\\tan(165) = -2+\sqrt{3}\\\\[/tex]

----------------------

As confirmation, you can use the idea that if x = y, then x-y = 0. We'll have x = tan(165) and y = -2+sqrt(3). When computing x-y, your calculator should get fairly close to 0, if not get 0 itself.

Or you can note how

[tex]\tan(165) \approx -0.267949\\\\-2+\sqrt{3} \approx -0.267949[/tex]

which helps us see that they are the same thing.

Further confirmation comes from WolframAlpha (see attached image). They decided to write the answer as [tex]\sqrt{3}-2[/tex] but it's the same as above.

Answer:

Option D

Only the equation in option D matches with the table

Answered by GAUTHMATH

Answer:

Solution given:

equation is:

x²-40x+A=0

when A=200

equation becomes

x²-40x+200=0

Comparing above equation with ax²+bx+c=0 we get

a=1

b=-40

c=200

x=[tex]\displaystyle \frac{-b±\sqrt{b²-4ac}}{2a}[/tex]

substituting value

x=[tex]\displaystyle \frac{-*-40±\sqrt{(-40)²-4*1*200}}{2*1}[/tex]

x=[tex]\displaystyle \frac{40±\sqrt{800}}{2}[/tex]

x=[tex]\displaystyle \frac{40±20\sqrt{2}}{2}[/tex]

taking positive

x=[tex]\displaystyle \frac{40+20\sqrt{2}}{2}[/tex]

x=34.14

taking negative

x=[tex]\displaystyle \frac{40-20\sqrt{2}}{2}[/tex]

x=5.86

Answer:

The area of circle is 12.56m sq.