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(a)

Let  a  = ( 0, m)

Let b = (m , 0)

Note that these are equal in length

(a + b)  =  (m , m)

(a - b)  =  (-m , m)

If  two vectors are perpendicular, their dot product  =  0

dot product =   (m , m)  ( dot )  (-m , m)  = -m^2  + m^2  =   0

Area =  L * B =     2B * B  = 2B^2

Yellow and Purple area =  (2/5 + 1/3) * 2B^2  =  (22/15)B^2

Pink area  = (1/2) ( 2B^2  - 22/15 *B^2 ) =  (4/15)B^2

So

Area - Painted Area =   240

2B^2 - (22/15 + (4/15))B^2  =  240

[4/15) B^2 =  240

B^2  = (240* 4 /15)  

B^2  = 960 / 15

B^2  =  64

B = 8

2B =  16

Perimeter  =  2 [ B + 2B ]  =   2 [ 8 + 16 ] =    48 cm

https://web2.0calc.com/questions/area_72

We know the three sides of PQR.....the other info doesn't matter

Semi-perimeter  =  [ 36 + 26+ 22 ]  /  2 =  84 / 2 =  42

Heron's Formula :

[ PQR ]  =  sqrt [ 42  * 6 * 16 * 20 ] =  48sqrt (35)   ≈ 283.97

B = (6/5)R

G = (3/2)B  →  G = (3/2)(6/5)R   =  (9/5)R

Total  number of shirts at te beginning  =  R + B + G  = R + (6/5)R  + (9/5)R =  4R

So  after he gives away  12 red shirts, we have

[R -12 ]  =  (1/10) 4R

R - 12 =  (2/5)R

R - (2/5)R  = 12

(3/5)R  =12

R =12 *(5/3)  =  20

The total shirts  he  had when he started =  4 (20)  = 80

After giving away 12 red shirts  he has  80  - 12  =   68   left

Since M is the midpoint of QR​, then QM=MR=2QR​=26​=3.

We now have a right triangle △PMQ with legs PQ=5 and QM=3. By the Pythagorean Theorem,

[PM^2 = PQ^2 + QM^2 = 5^2 + 3^2 = 25 + 9 = 34.]

Then, PM = sqrt(34).

Side of the square =  [product of leg lengths ] /  [ sum of leg lengths ] =  [ 1 * 3 ] /  [ 1 + 3] =   3 / 4

Area of square  =  [3/4]^2 =  9 / 16

Let the original cost  =  C

Let Adam's share = A

Adam would have paid    4 / (4 + 6) * C =   (4/10)C =  (2/5)C  =  A ....so....   C = (5/2)A 

The new cost is  20% more =   1.2C

So now Adam pays   1.2A = 57.60

A = 57.60 / 1.2 = 48

Original cost  

C = (5/2)A =  (5/2) *48 =  $120

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

Total ways: 35=24335=243. Ways with <3 yellow: 1+2(51)=211+2(15​)=21. So, 243−21=222243−21=222​ ways with at least three yellow squares.

Since M is the midpoint of QR​, then QM=MR=2QR​=26​=3.

Triangle PQR is a right triangle since ∠PQR=90∘ (by the Pythagorean Theorem, PQ2+QR2=PR2, so 52+62=72).

Since M is the midpoint of the hypotenuse of this right triangle, then segment PM is half the length of the hypotenuse. Therefore, PM = PR/2 = 7/2.

For this problem, you would have to set the amount of money as a variable. 

They key is to essentially write a solvable equation to figure it out!

In fact, my parents used to give me a lot of problems similar to this one :)

So let's let that be x. 

From the problem, we can get the equation \(\frac{2}{5}x+\frac{1}{7}(x-360)+360=x\). 

We have \(\frac{2}{5}x + \frac{1}{7}x - \frac{360}{7}+360 = x\). Combining like terms and isolating x, we have \(\frac{2160}{7}=\frac{16}{35}x\). 

We multiply both sides by 35/16, and we have \(\frac{35}{16} \cdot \frac{2160}{7} = x\). 

So we have \(x = 675\)

$675!

Thanks!

I don't think it is possible to achieve a real number from this function. 

This is because \(\sqrt{2}\) is greater than 1, so in order for it to be real, we would need \(\sqrt{3+\sqrt{x}}\) to be negative, which obviously is not possible...

I think the domain is essentially just \([0, \infty)\)

Thanks!

In order not to have two yellows next to each other,  

the three yellows must occupy squares 1, 3, and 5. 

That leaves squares 2 and 4 to fill, and two colors  

to do it with. There are only four ways. 

Blue Blue     Red Red     Blue Red     Red Blue  

So, the answer is four ways.  

_.

There are two events we need to consider:

Event 1: Drawing an even number first.

There are 4 colors in the deck.

Each color has one even number (2, 4, or 6).

So, there are 4 * 3 = 12 cards that satisfy this condition (even number, any color).

Event 2: Drawing a multiple of 3 after drawing an even number (without replacement).

After drawing the first card, there are only 27 cards remaining.

There are 3 multiples of 3 left (3 and 6, since one even number - 6 - is already drawn).

However, since Grok isn't replacing the first card, there are only 2 colors left that have multiples of 3 (red and blue, as the even number 6 was likely green or yellow).

Therefore, there are 2 * 1 = 2 cards that satisfy this condition (multiple of 3, remaining colors).

Total Favorable Cases:

To get the probability, we need the number of favorable cases (both events happening) divided by the total number of possible cases (drawing any two cards).

Favorable cases: 12 (Event 1) * 2 (Event 2) = 24

Total possible cases: There are 28 cards total (7 numbers * 4 colors), and we draw 2 without replacement. So, the total number of possible choices is 28C2 (28 choose 2) which is 28 * 27 / (2 * 1) = 378

Probability:

Therefore, the probability that Professor Grok draws an even number first, followed by a multiple of 3 (without replacement), is:

Probability = Favorable Cases / Total Possible Cases Probability = 24 / 378

Simplifying the fraction:

Both the numerator (24) and denominator (378) have a common divisor of 6. We can simplify:

Probability = (24 / 6) / (378 / 6) Probability = 4 / 63

So, the probability is 4/63.

We can solve this problem by using the properties of odd functions and the given information about f(x) and g(x).

Odd Function Property: An odd function satisfies f(-x) = -f(x) for all real numbers x. This means the function is symmetrical about the origin (0, 0).

Shifting the Graph: The function g(x) is defined as f(x + 3) - 5. This means the graph of g(x) is obtained by shifting the graph of f(x) three units to the left and five units down.

Point on the Graph of g(x): We are given that the graph of y = g(x) passes through the point (2, -2). This translates to f(2 + 3) - 5 = -2, which means f(5) = 3.

Odd Function and f(5): Since f(x) is odd, we know f(-5) = -f(5) = -3.

Points on the Original Graph (f(x)): Because g(x) is obtained by shifting f(x), the corresponding points on the graph of f(x) are:

Point for f(5): (-5, -3) (three units to the left of (2, -2) due to the shift)

Point for f(-5): (5, 3) (three units to the right of (2, -2) due to the shift)

Points on the Shifted Graph (g(x)): Since g(x) shifts the graph of f(x), the corresponding points on the graph of g(x) are:

Point for g(-5): (-2, -3) (three units to the left and five units down from (-5, -3))

Point for g(5): (8, 3) (three units to the right and five units down from (5, 3))

Therefore, the graph of y = g(x) must also pass through the points:

(-2, -3)

(8, 3)

So the answer is: (-2, -3), (8, 3)

Calculations: Given:

cos(ABD) = 4/5

sin(ABD) = 3/5

We can calculate:

sin(ABD / 2) = sin(IBD) = sqrt[(1 - 4/5) / 2] = sqrt(1/10)

cos(IBD) = sqrt[1 - 1/10] = 3/sqrt(10)

tan(IBD) = 1/3 = ID / 4

ID = 4/3

AD = 3 - (4/3) = 5/3

Triangles AMN and ABC are similar, and we have: AD / AI = (5/3) / 3 = 5/9

Calculating the areas of the triangles: [AMN] = (5/9)^2 [ABC] = (5/9)^2 (1/2) (8)(3) = (25 / 81 ) * 12 = 300 / 81 = 100 / 27

Answer: 100 / 27

https://web2.0calc.com/questions/triangle-hypotenuse-problem

W               A  2 X

                          b

                          C

                          3b

Z        B    24      Y

AC^2 =   b^2  +  2^2  =  b^2 + 4

BC^2  = 24^2 + 9b^2  =  9b^2 + 576 

AB^2  =  AC^2 + BC^2  =  10b^2  + 580

Also

AB^2  = (24 -2)^2  + (b + 3b)^2  =  484 +16b^2

So

16b^2 + 484 =  10b^2  + 580

6b^2  = 96

b^2  =  (96/6)  =  16

So

AB^2  = 10 (16)  + 580

AB^2  = 160 + 580

AB^2   =  740

AB = sqrt [740]  = 2sqrt (185)  ≈  27.2