All Answers 358355 Answers

D = 80, Q = 90, N = 40

2589 , 2598 , 2859 , 2895 , 2958 , 2985 , 3568 , 3586 , 3658 , 3685 , 3856 , 3865 , 4459 , 4495 , 4549 , 4566 , 4594 , 4656 , 4665 , 4945 , 4954 , 5289 , 5298 , 5368 , 5386 , 5449 , 5466 , 5494 , 5638 , 5646 , 5664 , 5683 , 5829 , 5836 , 5863 , 5892 , 5928 , 5944 , 5982 , 6358 , 6385 , 6456 , 6465 , 6538 , 6546 , 6564 , 6583 , 6645 , 6654 , 6835 , 6853 , 8259 , 8295 , 8356 , 8365 , 8529 , 8536 , 8563 , 8592 , 8635 , 8653 , 8925 , 8952 , 9258 , 9285 , 9445 , 9454 , 9528 , 9544 , 9582 , 9825 , 9852 , Total =  72

The first is the SMALLEST and the last is the LARGEST.

f(x) simplifies to x, so the range is all real numbers.

In interval notation, (-inf,inf).

f(10) - f(9) = 28.

Your number is: 1 577 5

2nd digit ==[1 + 7] / 2 + 1==5

3rd digit ==[5 + 7] / 2 + 1==7

4th digit ==[7 + 5] / 2 + 1==7

What is the slope of the line?

Hello Guest!

If the area of the triangle is 16, the line must intersect the y-axis at 4, because 8*4/2=16.

The slope of the line it 4 : 8 = 0.5

  !

Thank you!

Thanks!

45 = (1/2) (height) (2CD + CD)

90  =  (height) ( 3CD)

30  = (height) ( CD)

30  / CD  = height  of trapezoid

AMB  and BMC  are similar triangles

Since  AB = 2CD

Then height of  AMB  = 2 (height of  BMC)

So  there are three equal parts to the height of the trapezoid and  the height of AMB is two of them

So  height AMB    =  (2/3) (30) /( CD)  =  20 / CD

Area of AMB  = (1/2) (AB) [height (AMB) ]    =   (1/2) (2CD) [ 20/CD]  =  20

\(\frac{k(k-1)}{2}\) is really just the sum of integers from 1 to k-1, and the cosine of some integer multiple \(\pi\) oscillates between -1 and 1 (if the integer is even, it's 1, and if it's odd, it's -1). Here, the sum alternates 2 negatives and 2 positives (starting with k=19)

This means that the first 4 terms of the sum are

\(-(1+2+3+\cdots+18)-(1+2+3+\cdots+19)+(1+2+3+\cdots+20)+(1+2+3+\cdots+21) \\=20+19+20+21 \\=4(20)\)

(notice a bunch of cancellation happens between the sums, as the 1+2+3+...18 is repeated throughout all the 4 terms)

So for every group of 4 terms, the sum of the group is the biggest number of the sum of the 3rd term of that group times 4.

Therefore, the sum is:

\(4(20+24+28+\cdots+96)\)

Use the sum formula to obtain your answer.

Whoa nice!

f(0)  =  a/2

And   we can find  f^-1  as follows

y = a /(x + 2)     get x by itself

y ( x + 2)  = a

yx + 2y   = a

yx  =  a  - 2y

x =  (a - 2y) / y         swap x, y

y = (a - 2x) / x  = f^-1 (x)

f^-1(8a)  =   (a - 2(8a)) / (8a)  =  -15a /  (8a)  =   -15/8

So

a/2 =  -15/8

a = -15/4

Here's another way

cos B  =  sqrt (1 - sin^2B)  =  sqrt ( 1 - 25/49 )  =  sqrt ( 49 -25) / 7  =  sqrt ( 24)  / 7  =   

2sqrt (6)  /  7        (as BuilderBoi found !!!)

This is an unusual problem  !!!

See the graph here :

The  feasible area  =  (pi/2)^2  = pi^2/4

When x  = 0

sin (0)   - sin (y) + sin^2 (y)  =  3/4

1  - sin(y) +sin^2(y)  = 3/4

sIn^2 (y) - sin (y)  = -1/4

sin^2(y) - sin (y) + 1/4 = 0

(sin y - 1/2)^2  = 0

sin y  =  1/2

arcsin (1/2)  =  y

When  y =pi/2   we have (similarly)

sin^2 (x) -sin (x) + 1  =  3/4

sin^2 (x) -sin x = -1/4

sin^2 (x) - sin (x) + 1/4)  = 0

(sin x -1/2)^2 = 0

sin x  = 1/2

arcsin (1/2) = x

This sets  up  the area of  the two  right triangles  in the upper left and lower right of the feasible region

The area of these two right triangles  = ( pi/2 - arcsin (1/2)) ( arcsin (1/2))  

And the larger right triangle in the upper right corner has the area  (1/2) (pi/2- arcsin (1/2)^2

So  the area under consideration  =

2.4674  - .5483  - .5483 

pi^2 / 4   -  (pi/2 - arcsin (1/2))(arcsin (1/2))  - (1/2)(pi/2 -arcsin (1/2))^2  =   5pi^2  / 36

Here's my attempt (I'm not good at trig): 

C

5             7

A                           B

Let \(AC = 5\) and \(BC = 7\). 

By the Pythagorean Theorem, \(AB = \sqrt{24} = 2\sqrt6\)

So, \(\cos B = \color{brown}\boxed{2 \sqrt 6 \over 7}\)

We have the equation: \({2 - n \over n+1} + {2n - 4 \over 2 - n} = 1\)

Make a common denominator: \({(2 -n)(2-n) \over (n + 1)(2-n)} + {(2n-4)(n+1) \over (2-n)(n+1)} = 1\)

Simplify: \({n^2−4n+4 \over −n^2+n+2} + { 2n^2−2n−4\over -n^2 + n + 2} = 1\)

Add the fractions: \({3n^2 - 6n \over -n^2 + n + 2} = 1\)

Cross multiply: \(3n^2 - 6n = -n^2 + n + 2 \)

Move everything to the left-hand side: \(4n^2 - 7n - 2 = 0\)

Factor: \((4n+1)(n−2)=0\)

So, the solutions are \(-{1 \over 4} \) and \(2\).

But, to verify their validity, we need to plug them back into the original equation. 

Doing so, eliminates \(n = 2\), so \(n = \color{brown}\boxed{-{1 \over 4}}\)

Very nice, BB!!!!

1 - k(-3) = -3(-3) + 4 - 7(-3)

1 + 3k  = 9 + 4 + 21

1 + 3k =  34

3k =  33

k  = 11

We have: \({b+7 \over b + 4} = {c \over 9}\)

Cross multiply: \(9(b+7) = c(b + 4)\)

Distribute both sides: \(9b + 63 = bc + 4c \)

Subtract 9b from both sides: \(bc + 4c - 9b = 63\)

Factor out b: \(b(c-9) + 4c = 63\)

Subtract 36 from both sides (Simon's Favorite Factoring Trick): \(b(c-9) + 4c - 36 = 27\)

Factor out c: \(b(c-9) + 4(c-9) = 27\)

Simplify the left-hand side: \((b+4)(c-9) = 27\)

The (positive and negative) factor pairs of 27 are (1,27), (-1,-27), (-3,-9) and (3,9). 

Each of those cases has 2 pairs of integers that work, which makes for \(4 \times 2 = \color{brown}\boxed{8}\) integers. 

Thanks! I set up a venn diagram as you said and got the same answer!!

If B is in the second quadrant, the tangent of this angle must be  negative

it is 8

Assuming that everyone takes at least one of the subjects :

Number in School =   N(Chinese) + N(Spanish) - N (Both)

374  =  x  + 2x  - 100

474  = 3x

x = 474 /  3  =      158  = total taking Chinese

2x  = 2 (158)  = 316  = total taking Spanish

Proof  (you can set up in a Venn diagram)

58 take only Chinese   +   100 take both  +  216 take only Spanish

p - 2q  = 0    →   p = 2q             (1)

q - 2r  =  8  →    r  = (q - 8) / 2     (2)

p + r = 5     (3)

Sub (1) , (2) into (3)

2q  + (q - 8) / 2 =  5

4q + q - 8   =  10

5q  - 8  = 10

5q = 18

q = 18/5           p =  2(18/5) =  36/5         r  =  (18/5 - 8 ) / 2  =  -11/5

thank you!!! it was correct btw. :))