All Answers 355923 Answers

First one

a₁  = 4_n + 4

Second one

First term is 34

Each successive term is -8

So we have

a_n  =  42 - 8_n

First one :

The difference between successive terms is 9   and the first term is  21

So we must have that

a₁  = 12n + 9    ⇒   a₁  = 12 (1)  + 9   =  21

So...the last answer is correct

Second one

a_n  = 17  - 5n

The difference between successive terms is - 5   and the first term is 12

So we have

a_n  = a_n-1 - 5

a₁   = 12

Second answer

0 ≥ 54p - 144

0 > 12 - 20p

144 ≥ 54p    divide both sides by 54

144/54 ≥ p

8/3 ≥ p

p ≤ 8/3

0 > 12 - 20p

20p > 12    divide both sides by  20

p > 12/20

p > 3/5

The  answer is :

p  = (3/5, 8/3 ]

x  = 4cos t    ⇒  x/4  = cos t

y  = sin t

So...using  cos^2 t  +  sin^2 t  =  1, we have

(x/4)^2  + y^2   =  1

x^2 / 16  + y^2   =  1

This is an ellipse

When t  = 0   we have (4,0)

When t  = pi/2  we have  (0, 1)

When t  =  pi  we have (-4, 0)

When t  = 3pi/2  we have (0, -1)

When t   2pi  we have (4, 0)

So....the curve is traced out in a counter-clockwise direction

Here's the graph :  https://www.desmos.com/calculator/yscc2qxqsk

Here is a formula that will always work for these kind of problems

New Point  =  [ Old Point - Dilation Center ] * Scale Factor  + Dilation Center

Note that  (a, b) + (c - d)  = (a + c, b + d)

And that (a,b) - (c,b)  = (a - c, b - d)

Also  (a,b)*constant  =  (a*constant , b*constant )

So we have

New Point   =  [ (4,7) - (1,2) ] *10 + (1,2)  =

[(3, 5)]*10  + (1.2)  =

(30, 50)  + (1,2) =

(31, 52)

Use Geometric Series formula:

S = F x (1 - R^N) / (1 - R)

S = 6 x (1 - 2^6) / (1 - 2)

S = 6 x (1 - 64) / (-1)

S = 6 x ( -63) / (-1)

S = 6 x     63

S =378

What happened to your first question?

Since the train arrives randomly between 2:00 and 3:00 BUT WAITS FOR 20 MINUTES and then leaves. But when Lizzie also arrives randomly, she could luck out and arrive at that 20-minute window that the train is waiting. This means that Lizzie has 20 minutes out of 60 minutes, or 20 / 60 =1/3 probability that the train will be there when she arrives.

There are a total of $6^2$ or 36, different outcomes when you roll two die. 

If the blue die is x and the yellow die is y, their outcome will be (x,y)

We set x > y, so we have:

(6,1) ; (6,2) ; (6,3) ; (6,4) ; (6,5) ; (5,4) ; (5,3) ; (5,2) ; (5,1) ; (4,3) ; (4,2) ; (4,1) ; (3,2) ; (3,1); and (2,1).

15 out of the 36 outcomes work. 

The answer you are looking for is 15. 

The strategy here is see how many numbers there are, and then subtract the total amount of perfect squares.

There are a total of $250 - 50 + 1$, or 201, numbers.

Between 50 and 250, there are perfect squares:

64, 81, 100, 121, 144, 169, 196, and 225.

8 out of the 201 numbers are perfect squares.

So there are 201 - 8 numbers that are not perfect squares. 

‘

Your final answer will be 193. 

How many numbers between 1 and 1,000 are divisible by 9?

1,000 / 9 = 111 numbers that are multiples of 9 and thereby divisible by 9.

1,000 - 111 =889 numbers that are not divisible by 9.

Alright, here we go:

Since all we are talking about is heads or tails, what type of coin it is doesn’t matter.

There are a total of $2^3$ different outcomes when you flip three coins. 

Here are all of them listed:

HHH ; HHT ; HTH ; THH ; TTT ; TTH ; THT ; HTT

Out of the eight different outcomes, only $\binom{3}{2}$ of them have 2 heads.

HHT, THH, and HTH

Since the problem is asking for at least, we can also get 3 heads, which there is only one way to do: 

HHH.

There are 4 outcomes that match what the problem is asking for, so the answer is 4/8, 0.5, or 50%. 

It's the max error that is linear with h:

.

You can’t you need to solve tan then continue

A further thought.

Repeat your calculations for x = 0 to x = pi/2 and see if the errors behave as they should.

Tiggsy

I see, thank you very much 

It's been a while since I dealt with this sort of stuff, but I think that this is the likely explanation.

For the ode 

\(\displaystyle y'=f(x,y) ,\)

the local truncation error is

\(\displaystyle \frac{h^{2}}{2!}f'(x_{0},y_{0}),\)

which, for the given equation will be

\(\displaystyle \frac{h^{2}}{2!}\cos(x_{0}).\)

Since we are working over the range 0 to pi, what will happen is that the errors to the left of pi/2 will, in some way, cancel out with the errors to the right of pi/2, (because the cosines will be equal in magnitude but opposite in sign).

That means that the truncation errors don't build up in the usual way. Looking at the truncation errors for each step will probably give a clearer picture.

Tiggsy

3)

Outcomes of equal chance

Even even

Even odd 

Odd even

Odd odd

Only the 2 odds wont multiple to an even so that's 3/4 chance that the product will be even.

No of ways = 100×100-50×50 = 7500

The probability that 2 babies will speak tomorrow is C(5,2) · (1/4)² · (3/4)³

                                3                                                C(5,3) · (1/4)³ · (3/4)²

                                4                                                C(5,4) · (1/4)⁴ · (3/4)¹

                                5                                                C(5,5) · (1/4)⁵ · (3/4)⁰

Find the answer to each of the above and add them together.

The line     y = (2/3)x     intersects the line     y = 4     at the point     (6, 4).

Therefore, the sum of the coordinates of that point is 10.

Call the intersection of BE with CH point X. Then the triangle BCX is a right isosceles right triangle.

This means that angle CBE is a 45 degree angle. 

And, both BX and CX are equal to 6·sqrt(2).

CX is a height of the trapezoid and equals 6·sqrt(2).

CD is one base of the trapezoid and equals 12.

BE is the other base of the trapezoid and equals 6·sqrt(2) + 12 + 6·sqrt(2) = 12 + 12·sqrt(2).

The area of the trapezoid is (1/2) · height · (base #1 + base #2)

                                       =    (1/2) · 6·sqrt(2) · ( 12 + 12·sqrt(2) )

                                       =    3 ·sqrt(2) · ( 12 + 12·sqrt(2) )

                                       =  36·sqrt(2) + 72

1)  (f + g)(x)  =  (x³ - x) + (x³ + 2x² - 10)  =  (x³ + x³) + 2x² - x - 10  =  2x³ + 2x² - x - 10

2)  (f * g)(x)  =  (x² + 2x - 6)(x⁴ + 3)  =  x⁶ + 2x⁵ - 6x⁴ + 3x² + 6x -18)                    [distributive property]

3)  (f - g)(x)  =  (x⁴ - x² + 9) - (x³ + 3x² + 12)  =  x⁴ - x² + 9 - x³ - 3x² - 12  =  x⁴ - x³ - 4x² - 3

5^2+12^2=13^2 so the segment is 13 units

X=19

4 because the second equation is the first one with each term double so 2*2=4