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41. Here is the solution:

Imagine the unit circle: 

We know that 300 degrees is in the fourth quadrant, where cosine is positive.

300 degrees has a reference angle of 60 degrees, since it is 60 degrees away from the x-axis

Since \(\cos{60º}=\frac12, \cos{300º}=\frac12\)

I hope this helped,

Gavin

42. ?

41. cos 300=1/2

1st one is true, 20-0=20

20-0=20

1. First,  find the mean: 87.2

2. We find the distance from each data point to the mean, 

First point: 83- square distance from mean- 17.64

85-87.2-4.84

82 - 87.2 -27.04

93-87.2 squared-33.64

83-87.2- 17.64

84-87.2-10.24

95-87.2-60.84

87-87.2-0.04

86-87.2-1.44

94-87.2- 46.24

Now, add everything up: 201.96

10 data points, so 201.96/10=20.196

Take the square root of that to get:\(\sqrt{20.196}=4.49..\)

So, 4.6?

Fiestygeco, if you are arrested and imprisoned for this, I’ll bribe the guards and have them smuggle in some juicy bugs for you. June Bugs and May Beetles are my favorite—and they are in season now.

Yours truly,

The Salamander

Hey RB!

I'm not that good at number theory but here is what I have:

Multiplying both sides of the congruence:

\(2x\equiv y+5\pmod 9 \)

by 5 gives:

\(10x \equiv 5y+25\pmod 9,\)

then reducing both sides modulo 9 gives

\( x\equiv 5y+7\pmod 9. \)

Thus, the product of the blanks is \(5\cdot 7=\boxed{35}.\)

I hope this helped,

Gavin

Humanities-male +female=150

Male for Humantities-70

\(\frac{70}{150}\approx0.467\)

The Answer is 168.

L =Lev, M=Mina, N=Naomi

L/M = 1/2 and M/N=3/4. Therefore:

L=3N/8, M=3N/4, N = (4 M)/3. So if Lev is 1-year old, then:

L= 1, M= 2, N =2 2/3. So the ratio of:

L : M : N =1 : 2 : 2 2/3

Thanks everybody so much

Although the notation is sloppy, I will assume based on former questions of the same that "∑n=15[5(−2)^n−1]" translates mathematically to \(\sum_{n=1}^5[5(-2)^{n-1}]\). Please correct me, though, if this in incorrect. 

Anyway, there is neat formula that will deal with question very quickly. It is the following:

\(S_n=a_1\left(\frac{1-r^n}{1-r}\right),r\neq1\)

S_n is the sum of the geometric series

a₁ is the first term in the series

r is the common ratio

n is the number of terms in the finite geometric series

Let's use this formula to our advantage! However, there are two variables that we do not know: a₁ and  n. Let's start with a₁.The first term in the series is when n=1, so let's determine what that is. 
 

The number of terms also must be calculated:

You realize that these are copyrighted problems, right? It literally says "Copyright © AoPS Incorporated. This page is copyrighted material. You can view and print this page for your own use, but you cannot share the contents of this page with others" on the site where these homework problems are.

Yes, you're almost right, it's because 4 + 5 is less than 10. No matter which direction the forces are acting in, the 10 N force will be greater than any opposing force, so the system cannot be in equilibrium.

This is the sum of an arithmetic series

The first term  is   3(1)  - 1  = 2

The eighth term is 3(8)  - 1 = 23

The sum is

[first term + eigth term]  * number of terms  / 2  =

[ 2 + 23  ] * 8  / 2  =

[25] * 4  =

100

This  will be the sum of a geometric series where the first term is 4

The common ratio  is -5

And we will have 6 terms.....so....the sum is

first term   [ 1  -  common ratio^number of terms ] / [ 1 - common ratio ]  =

4  [1 - (-5)^6]  /  [ 1  - -5 ]  =

4 [ 1 - 15625 ] / 6  =

(2/3) ( -15624 )  =

-10416

Sketch the parabola x2 = 4by . The tangent and normal at a point P( 2bt, bt2)

meet the axis

of the parabola in D, E respectively.
(i) Find the coordinates of D and E.
(ii) Show that the focus Q is the midpoint of DE.
(iii) Hence prove that Q is the centre of a circle through D, P, E.
(iv) If the normal at P intersects the x-axis in F find F and the length of PF. Hence prove that PF:PD = t:2

a_n  = 7n  - 2

a_n   =1500 (1.15)^{n -1}   ⇒ last answer

The  amount of chocolate  contained  in the large hemisphere  is just

(2/3)pi (1)^3  =  (2/3) pi ft^3

So....we want to solve this for r  to find the radius of each of the smaller hemispheres

(2/3) pi   =  27  (2/3) pi (r)^3

1  = 27 * r^3

1/27 =  r^3

1/3  = r   

So....r = 1/3 ft  

This seems to be harder than it really is......

Notice that if we "unwrapped"  one  complete  revolution  of the string around the post, we would have a rectangle with a "width"  of  4 ft  [the circumference of the post ].....and the height of this rectangle would be just 12 / 4  =  3ft

And the string would run  from the top left of this rectangle to the bottom right, thus comprising the diagonal of this rectangle .......so....the length of this string would be the hypotenuse of a right triangle with legs of 3 and 4.....and its length  =  sqrt  [3^2 + 4^2]  = sqrt [ 25]  = 5

And we have 3 more  similar rectangles in the next 3 revolutions.....so....the string length  is just

1 revolution * 5 ft   +   3 revoultions * 5 ft  = 

4 * 5 ft  =

20 ft

18

Note that the  first term  is 2(1) + 1 =  3   and the last term  is 2(12) + 1  = 25

So...the sum is

[ first term + last term ]  *  [ number of terms ] / 2  =

[ 3 + 25 ] *  [ 12]  / 2  =

28 *  6  =

168

17

5th term  = 140

7th term = 35

We have that  

(140)r^2  = (35)       where r  is the common ratio

Divide both sides by  140

r^2  = 35/140  =  1/4        take both roots

r = +/-  sqrt  (1/4)   =  +/- 1/2

So....note that the 6th term could be  (140)(1/2)  = 70

Or.....the 6th term could be (140) (-1/2)  =  -70

Thus.....  ± 70   is correct

Thank you!