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it's 10472.4ft

1.  In the form  (x - h)^2  + (y - k)^2  = r^2, the center is  (h, k)

So we have

(x + 7)^2  +  (y - 13)^2  =  196

(x  -  (-7) )^2  + (y - 13)^2  = 196

So.....the center  is   (-7, 13)

2. x^2 + y^2  + 14x + 10y - 7  = 0        re-write as

x^2 + 14x  +  y^2  + 10y     =   7        complete the square on x and y

x^2 + 14x + 49  + y^2  + 10y + 25  =  7 + 49 + 25     factor and simplify

(x + 7)^2  + (y + 5)^2  =  81

So....the center is  (-7, -5)

3.  Hyperbola

Let the height of the plane above the helicopter be "h"

We have a right triangle  where  the side opposite the 71 degree angle  =  h   and the side adjacent to this angle =  3055

So   we can use the tangent here

tan (71)  =  h / 3055       multiply both sides by 3055

3055 * tan (71)  =  h ≈   8872.4  ft

So.......the total altitude of the plane is

1600 + 8872.4  =

10472. 4 ft

No. This is a variant of the Chicken McNugget theorem. It’s 7x+10y+53z=d. We want to find the biggest value of d where it cannot be expressed as the left hand side. Also, x, y, z, are all positive. I personally dunno how to do this type thoe. Somebody else maybe post solution? C Phill? Melody?

1. The 12 ft   can be considered the hypotenuse of a right triangle

And the 8 ft  can be considered the leg opposite the angle we are considering

So....to  find out if this angle is at least 45°, we have take thw sine inverse of  8/12

So  we have

asin (8/12)  =  the angle  ≈ 41.8°.....so....it won't pour out smoothly

2. Area  =  (3.14) r^2 *  (40/ 360)  =

(3.14) * 4^2 * 1/9   =   

16 * 3.14 / 9    =

5.6 cm^2

3. We have  a 30-60-90  right triangle.....the side opposite the 30 degree angle  = 10 = base

And the side oposite the 60 degree angle  = 10√3 cm = height

So....the area  is   (1/2) base * height  =

(1/2) 10 * 10√3  = 

50√3   cm^2  ≈    86.6 cm^2

1)

This is the formula you would use to figure that out:

FV = PV [1 + R]^N

FV = 24,000 [1 + 0.08]^3

FV = 24,000 [1.08]^3

FV = 24,000 x 1.259712

FV =$30,233.09 - This is the value of investment in 3 years.

Use exact same formula above to calculate the rest of them.

=$38,084.98 - This is the value after 6 years.

=$47,976.11 - This is the value after 9 years.

=$60,436.08 - This is the value after 12 years.

2)

This is the formula you would use to calculate that:

PV = FV / [1 + R]^N

PV =120,000 / [1 + 0.03]^5

PV =120,000 / [1.03]^5

PV =120,000 /  1.1592740743

PV =$103,513.05 - This is the PV of $120,000 @ 3% over 5 years.

Use exact same formula above to calculate the rest of them:

=$89,670.98 - This is the PV @ 6% over 5 years.

=$77,991.77 - This is the PV @ 9% over 5 years.

=$68,091.22 - This is the PV @12% over 5 years.

1.  We can find AE  with the secant product theorem

CA * AB   =  AE * DE

18 * 3   =  AE * AD

54  =  AE * AD

Let  AD  =  x

Then AE  =  3 + x        and we have that

54  = (3 + x)* x

54  = 3x + x^2       rearrange as

x^2 + 3x - 54   =  0        factor

(x + 9) (x - 6)  = 0

Setting each factor to 0 and solving for x and we have that

x  = -9    (reject)          and      x  = 6  (accept)

So

x = AD  =  6

And AE  =  3 + x   =   3 + 6   =  9

I'm assuming that the books are arranged on separate shelves, so that Diane knows exactly what books she is choosing. If that is correct, then she can choose in: 12C3 = 220 different ways of choosing 3 hardcover detective novels.

Guest stop being rude. You make rude comments all the time so just stop. Nobody likes this so just STOP.

Take the integral:

integral(7 x)/2 dx

Factor out constants:

= 7/2 integral x dx

The integral of x is x^2/2:

= (7 x^2)/4 + constant

Which is equal to:

= 1.75x^2 + constant

Well, if we are allowed to add AND subtract the 3 coins, then I cannot think of not being able to make up ANY number by finding the proper values of a, b, c. Now, for exmple, you would think that it would be difficult or not possible to make up a number 1 greater than the LCM of 7, 10, 53. But, that is not the case because: LCM [7, 10, 53] =3,710 + 1 =3,711. I can easily make up 3,711 as follows:

[69 x 53] + [ 7 x 2] + [ 10 x 4] =3,711........and so on and in many different combinations of the 3 coins.

Well, If he has 4 kinds of flowers, then he has 4 choices for the first quadrant, 3 choices for the second quadrant, 2 choices for the third quadrant and finally 1 choice for the fourth quadrant(this is based on his desire not to have adjacent quadrants planted with same kind of flower.)

So: 4 x 3 x 2 x 1 = 24 different gardens of flowers that he could plant.

Note: I didn't understand what you meant by "the gardner can use the same flower more than once" Did you mean within a particular quadrant, or throughout the four quadrants?

This is how i understood his question (correct me if i'm wrong):

Find largest natural number that we can't express as 7*a+10*b+53*c where a, b, c are non-negative integers.

See the answer here:

https://web2.0calc.com/questions/time_5

If I understand your question, you want to know if there is a maximum amount that you CANNOT make by adding or subtracting these 3 coins. If that is what is meant, then I think there is NO MAXIMUM that you cannot make. My reasoning is that you can make ALL the numbers from 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 and up just from the 7 and 10 by adding and subtracting them. Example: If you want to make 1,000,000 from these 3 coins, you could easily do it this way: [18,866 x 53] + [6 x 7] + [6 x 10] = 1,000,000.

Let us know if that is what is meant by the question.

Nvm there are real solutions to this equation.

Sorry!

Let $f(x) = x + 2$ and $g(x) = 1/f(x)$. What is $g(f(-3))$?

\(\begin{array}{|rcll|} \hline f(x) &=& x + 2 \\ f(-3) &=& -3 + 2 \\ &=& -1 \\\\ g(x) &=& \frac{1}{f(x)} \\ g(f(-3)) &=& g(-1) \\ &=& \frac{1}{f(-1)} \quad & | \quad f(-1) = -1 + 2 = 1 \\ &=& \frac{1}{1} \\ &=& 1 \\\\ \mathbf{g(f(-3))} &\mathbf{=}& \mathbf{1} \\ \hline \end{array}\)

Did I do it correct?

\(\begin{array}{|rcll|} \hline \vec{x} &=& \vec{i} + 2\vec{k} \\ \vec{y} &=& 3\vec{i} + \vec{j} \\ \vec{z} &=& 3\vec{j} + \vec{k} \\\\ && \mathbf{2(\vec{x}-\vec{y})+\vec{z}} \\ &=& 2\left(\vec{i} + 2\vec{k}-(3\vec{i} + \vec{j}) \right)+3\vec{j} + \vec{k} \\ &=& 2\left(\vec{i} + 2\vec{k}-3\vec{i} - \vec{j} \right)+3\vec{j} + \vec{k} \\ &=& 2\left( -2\vec{i} - \vec{j} +2\vec{k} \right)+3\vec{j} + \vec{k} \\ &=& -4\vec{i} - 2\vec{j} +4\vec{k} +3\vec{j} + \vec{k} \\ &\mathbf{=}& \mathbf{-4\vec{i} +\vec{j} + 5\vec{k}} \\ \hline \end{array}\)

Let $f(x) = x + 2$ and $g(x) = 1/f(x)$. What is $g(f(-3))$?

1. Did you mean (a-x)^1/2=y?

2. Do x and a have to be real? (There are no real solutions to this equation)

Thank you so much for all your help, Heureka!!!

Does the question let you use a calculator? If so, you simply just enter each question on the calculator (by pressing the appropriate button (i.e sin, cos, tan) that is relevant) and right down the answers provided.

These would be the results:

sin(20) = 0.912945250728

cos(80) = -0.110387243839

tan(35) = 0.473814720414

sin(51) = 0.670229175843

cos(17) = -0.275163338052

I hope this helps.

Let $f(x) = Ax + B$ and $g(x) = Bx + A$, where $A \neq B$. If $f(g(x)) - g(f(x)) = B - A$, what is $A + B$?

\(\begin{array}{|lrcll|} \hline f(x) = Ax + B \\ g(x) = Bx + A \\\\ f(g(x)) = f(Bx+A) = A(Bx+A)+B \\ g(f(x)) = g(Ax+B) = B(Ax+B)+A \\\\ f(g(x)) - g(f(x)) = A(Bx+A)+B - [B(Ax+B)+A] &=& B-A \\\\ \begin{array}{rcll} A(Bx+A)+B - [B(Ax+B)+A] &=& B-A \\ A(Bx+A)+B - B(Ax+B)-A &=& B-A \\ A(Bx+A) - B(Ax+B) &=& 0 \\ A(Bx+A) &=& B(Ax+B) \\ ABx+A^2 &=& BAx+B^2 \\ A^2 &=& B^2 \\ A^2-B^2 &=& 0\\ (A-B)(A+B) &=& 0 \\ A+B &=& \dfrac{0}{A-B} \quad | \quad A \neq B \ !\\ \mathbf{ A+B } & \mathbf{=} & \mathbf{0} \\ \end{array} \\ \hline \end{array} \)

Vector Problem

\(\begin{array}{|rcll|} \hline && |3\vec{x}+2\vec{y}| \qquad \text{Formula: $|\vec{v}|=\sqrt{\vec{v}\cdot \vec{v}}$ } \\\\ &=& \sqrt{ (3\vec{x}+2\vec{y}) (3\vec{x}+2\vec{y}) } \\ &=& \sqrt{ 9\vec{x}^2 + 4 \vec{y}^2 + 2\cdot 3 \cdot 2\cdot( \vec{x}\cdot \vec{y}) } \\\\ && \qquad \text{Formula: $|\vec{x}^2|=|\vec{x}|^2 = x^2 \qquad |\vec{y}^2|=|\vec{y}|^2 = y^2$ } \\\\ &=& \sqrt{ 9x^2 + 4 y^2 + 12 \cdot(\vec{x}\cdot \vec{y}) } \\\\ && \qquad \text{unit vectors: $|\vec{x}|=x=1 \quad|\vec{y}|=y=1$} \\\\ &=& \sqrt{ 9 + 4 + 12 \cdot (\vec{x}\cdot \vec{y}) } \\\\ && \qquad \text{Formula: $\vec{x}\cdot \vec{y} = |\vec{x}|\cdot |\vec{y}| \cdot \cos(120^{\circ})=1\cdot 1\cdot (-0.5)=-0.5 $ } \\\\ &=& \sqrt{ 9 + 4 + 12 \cdot (-0.5) } \\\\ &=& \sqrt{ 13 -6 } \\\\ &\mathbf{=}& \mathbf{\sqrt{ 7 }} \\ \hline \end{array} \)