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I got 25/cos13

The 13 degrees is between the anchor rope and the verticial depth.

So you have the hypotenuse and the adjacent sides.

Thank you NinjaDevo,  

That is a really nice offer.

nice one ninja and thanks for the offer ! 

(1)

      4 * cars + 2 * bicycle = 110 | : 2

      2 * cars +       bicycle = 55

(2)  

            cars +       bicycle = 40

(1) - (2)

       2 * cars - cars + bicycle - bicycle = 55 - 40

       cars + 0 = 15

       cars = 15

       bicycle = 40 - cars = 40 - 15 = 25

cars = 15 and bicycle = 25

math or maths is short for mathematics.

$$\\\boxed{\vec{v}*\vec{u} = \begin{pmatrix} \textcolor[rgb]{1,0,0}{v_1}\\ \textcolor[rgb]{0,0,1}{v_2} \end{pmatrix} * \begin{pmatrix} \textcolor[rgb]{1,0,0}{u_1}\\ \textcolor[rgb]{0,0,1}{u_2} \end{pmatrix} = \begin{pmatrix} \textcolor[rgb]{1,0,0}{v_1}*\textcolor[rgb]{1,0,0}{u_1} + \textcolor[rgb]{0,0,1}{v_2}*\textcolor[rgb]{0,0,1}{u_2} \end{pmatrix}}\\
\boxed{\vec{v}*\vec{u} = \begin{pmatrix} \textcolor[rgb]{1,0,0}{3}\\ \textcolor[rgb]{0,0,1}{8} \end{pmatrix} * \begin{pmatrix} \textcolor[rgb]{1,0,0}{2}\\ \textcolor[rgb]{0,0,1}{-6} \end{pmatrix} = \begin{pmatrix} \textcolor[rgb]{1,0,0}{3}*\textcolor[rgb]{1,0,0}{2} + \textcolor[rgb]{0,0,1}{8}*\textcolor[rgb]{0,0,1}{(-6)} \end{pmatrix}=\textcolor[rgb]{1,0,0}{6}\textcolor[rgb]{0,0,1}{-48}=-42}$$

This is wrong: $$\sqrt{(3*2)^2+(8*(-6))^2} = 48.374 \;?!$$

first drag 10^-9 to the top since its the same as 10^9

so .09*10^9/3*10^4*2

then get rid of 10^4 on the bottom by removing 4 from the 10^9 on top

so. .09*10^5/3*2

now just multiply .09 and 10^5 =9000 on top and 3*2 to 6 on bot

so 9000/6 and then divide to 1500

The answer is 1500

This has to be an angle in the 3rd quagrant for the secant to be negative and the tangent to be positive.

The secant is = 2 where the cosine = 1/2, i.e., 60 degrees. So this is a 60 degree angle in quad 3 = 240 degrees.

The sine = -√(3)/2   and the cotangent = 1/√(3) (or √(3)/3)

A = (b)(h)/2   and since the base is 2 times the height, we can use

A = (2h)(h)/2 =  h²

9 = h²    ....so, it appears that h= 3   (the base is twice that = 6)

If f(x) = -10x  ...  then let g(x) be the same function with a shift right of 5 units...then

g(x) = -10(x - 5)

See the graph below.....the red line is f(x).....the blue is g(x)

Right, CPhill! That does make more sense for this problem:)

kitty<3...I like your answer, but I might suggest one small change.... how about this??

h(t) = -15t + 372  .... where "h" is some height after "t" seconds

Using "x" makes it seem as if there's some "horizontal" component to the problem...

Does this sound reasonable ??  (Who knows...I could be crazy !!)

See the picture....Sean is at "A'  about 287 ft off the ground  ....LeAnn is at "B" ...her "horizontal" distance from Sean (CB) is about 410 ft.....the line of sight distance from Sean to LeAnn (AB) is 500 ft.

From the given information, you already know one point on the line, as well as the slope. 

Point: (0, 372)

Slope: -15

You can now find the function by using the point-slope formula: y-y₁ = m(x-x₁)

Plug in the point and the slope:

y - 372 = -15 (x-0)

Add 372 to both sides.

y = -15x + 372

So f(x) = -15x + 372

haha bottom ok first take the derrivative of the quatritac formula (ax2+bx+c) and multiply by the tangent of the parabolic curve. 

so since i am looking for the adj side the hyp is 500 adj is x and 35 degrees is like him leaning to see her

No.... the tanget can't be used because we don't know the opposite side (the horizontal distance between Sean and LeAnn if Sean were standing on the ground) or the adjacent side (Sean's height off the ground)

I thought it would be tan?

So..if you can imagine what we have...The line of sight distance is the hypoteneuse of a right triange. Sean's height off the ground is the adjacent side of an angle that measures (90 - 35) = 55 degrees. So we have

cos(55) = x / 500     where "x" is Sean's height off the ground

500cos(55) = x = about 287 ft.

I can provide a picture.....if you need it.....

$${\frac{{\mathtt{1}}}{{{\mathtt{3}}}^{{\mathtt{2}}}}}{\mathtt{\,\times\,}}{\mathtt{6}}$$

$${\frac{{\mathtt{1}}}{{\mathtt{9}}}}{\mathtt{\,\times\,}}{\mathtt{6}}$$

$${\frac{{\mathtt{6}}}{{\mathtt{9}}}}$$

$${\frac{{\mathtt{2}}}{{\mathtt{3}}}}$$

So...we know the adjacent side and we're looking for the opposite side...looks like we need to use the tangent function here...so we have

tan (32) = (x-2.2) / 15    ......note that Bob's line of sight is 2.2 m "up" the pole, so (x-2.2) will represent the rest of the pole length with "x" being the height  ... so we have

15 tan(32) = x - 2.2       add 2.2 to each side

2.2 + 15 tan(32) = x = 11.57 m

Remember that if we have u = < u₁ , u₂ > and v = < v₁ , v₂ > .... the dot product is just a scalar given by:

(u₁* v₁) +  (u₂* v₂)  =  dot product

Try it again.......there shouldn't be any decimals in this answer....repost if you still have trouble.......

Correct answer.....

Let's "clean it up" a little

v = <5,7>

llvll = √(5² + 7²) = √(25 + 49) = √(74) ≈ 8.6

A little better, huh???

20x - 10 = 14x + 8

20x = 14x + 18

(20-14)x = 18

6x = 18

x = 18/6

x = 3