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Recall the compound angle formula: 

\(\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B\\ \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B\)

We then have:

\(\quad \cos C\\ = \cos(180^\circ - A - B)\\ = -\cos(A + B)\\ =-\cos A \cos B + \sin A \sin B\\ =-\sqrt{\dfrac7{10}} \sqrt{\dfrac{3}{10}} + \sqrt{1 - \left(\sqrt{\dfrac7{10}}\right)^2} \cdot \sqrt{1 - \left(\sqrt{\dfrac3{10}}\right)^2}\\ =-\sqrt{\dfrac7{10}} \sqrt{\dfrac{3}{10}}+\sqrt{\dfrac7{10}} \sqrt{\dfrac{3}{10}}\\ = 0\)

This shows that \(\triangle ABC\) is a right triangle with \(\angle C = 90^\circ\).

We can find cos(C) using the Law of Cosines. Here's how:

Law of Cosines:

The Law of Cosines relates the sides and angles of a triangle. For triangle ABC, it states:

c^2 = a^2 + b^2 - 2ab * cos(C)

where:

c is the side opposite angle C (in this case, side AB)

a is the side adjacent to angle A (in this case, side BC)

b is the side adjacent to angle B (in this case, side AC)

Given Information:

We are given:

cos(A) = sqrt(7/10)

cos(B) = sqrt(3/10)

We need to find cos(C).

Sides are Unknown:

We don't have information about the side lengths (a, b, or c). However, we can express them using cos(A) and cos(B) for further calculations.

Expressing Sides:

Using the trigonometric identity cos^2(x) + sin^2(x) = 1, we can find sin^2(A) and sin^2(B):

sin^2(A) = 1 - cos^2(A) = 1 - (sqrt(7/10))^2 = 3/10

sin^2(B) = 1 - cos^2(B) = 1 - (sqrt(3/10))^2 = 7/10

Relating Sides to Cosines:

Since we are looking for cos(C), let's focus on side c (AB). Using cosine definitions in right triangles:

cos(A) = (adjacent side) / (hypotenuse) = a / c

cos(B) = (adjacent side) / (hypotenuse) = b / c

Therefore:

a = c * cos(A) = c * sqrt(7/10)

b = c * cos(B) = c * sqrt(3/10)

Applying the Law of Cosines:

Substitute the expressions for a and b in the Law of Cosines equation:

c^2 = (c * sqrt(7/10))^2 + (c * sqrt(3/10))^2 - 2 * c * sqrt(7/10) * c * sqrt(3/10) * cos(C)

Simplifying the equation:

c^2 = (7/10)c^2 + (3/10)c^2 - (6/10)c^2 * cos(C)

Combining like terms:

c^2 = (7 + 3 - 6) / 10 * c^2 - (6/10)c^2 * cos(C)

4c^2 / 10 = -(6/10)c^2 * cos(C)

Solving for cos(C):

Isolate cos(C):

cos(C) = -(4c^2 / 10) / -(6/10)c^2

cos(C) = 4/6 (Since both c^2 terms are negative, the negative signs cancel out)

cos(C) = 2/3

Therefore, cos(C) = 2/3.

To solve this problem, let's first find the total number of possible pairs of cards that Professor Grok can draw. Since there are 4 colors and 7 numbers, there are \(4 \times 7 = 28\) cards in total.

Now, let's find the number of pairs where the first card drawn is labeled with an even number and the second card drawn is labeled with a multiple of 3.

1. There are 7 even numbers among 1 to 7: 2, 4, 6.

2. There are 7 multiples of 3 among 1 to 7: 3, 6.

To find the number of pairs with the desired properties:

- For the first card, there are 7 even numbers, and for the second card, there are 7 multiples of 3.

- So, the total number of pairs with the desired properties is \(7 \times 7 = 49\).

Now, to find the probability, divide the number of pairs with the desired properties by the total number of possible pairs:

\[\text{Probability} = \frac{\text{Number of pairs with desired properties}}{\text{Total number of possible pairs}} = \frac{49}{28 \times 27}\]

However, since Professor Grok is drawing without replacement, after the first card is drawn, there are only 27 cards left. So, the denominator should be \(28 \times 27\). 

Thus, the probability that the first card Grok draws is labeled with an even number, and the second card Grok draws is labeled with a multiple of 3 is:

\[\frac{49}{28 \times 27}  = \frac{7}{108}.\]

Let's analyze each statement based on the given conditions:

a) a + c > b + d: Since a > b and c > d, adding them together will preserve the inequality. So, a + c > b + d is MUST be true.

b) 2a + 3c > 2b + 3d: Multiplying both sides of a > b and c > d by positive constants 2 and 3, respectively, will still hold true. So, 2a + 3c > 2b + 3d MUST be true.

c) a - c > b - d: Subtracting c from both sides of a > b doesn't necessarily guarantee a - c > b - d. It depends on the relative values of a, b, and c. Not necessarily true.

d) ac > bd: Since a > b and c > d, multiplying them together will maintain the inequality as long as both a and c have the same sign (both positive or both negative). This is MUST be true.

e) a^2 + c^2 > b^2 + d^2: Squaring an inequality doesn't necessarily preserve the direction of the inequality. It depends on the signs of a and b. Not necessarily true.

f) a^3 + c^3 > b^3 + d^3: Similar to case (e), cubing an inequality doesn't guarantee the same direction for the result. Not necessarily true.

g) a^100 + b^100 > c^100 + d^100: The behavior of high-power exponents is unpredictable with inequalities. We cannot determine the direction of the inequality based on the given information. Not necessarily true.

h) a + b > c + d: Since a > b and c > d, adding b to both sides of a > b doesn't necessarily make the sum larger than c + d. It depends on the relative values of a, b, c, and d. Not necessarily true.

Summary: The statements that MUST be true based on the given conditions are:

a) a + c > b + d

b) 2a + 3c > 2b + 3d

d) ac > bd

Therefore, the answer is a, b, and d.

This expression is a nested fraction. We can simplify it by working our way from the innermost level outward.

Innermost Level:

21​=0.5

Second Level:

4+0.51​=4.51​=92​

Third Level:

4+92​1​=936+2​1​=938​1​=389​

Outermost Level:

1+389​1​=3838+9​1​=3847​1​=4738​​

Therefore, the value of the expression is 4738​.

Let's solve the equation for c by considering the absolute value signs. There are three cases to analyze depending on the sign of c + 5.

Case 1: c + 5 ≥ 0

In this case, the absolute value simplifies to c + 5. The equation becomes:

(c + 5) - 3c = 10 + 2(c - 4) - 6c

Simplifying:

-2c + 5 = 10 - 4c - 8 2c = -3 c = -\dfrac{3}{2} (Not a solution because c + 5 < 0)

Case 2: -5 ≤ c + 5 < 0

Here, the absolute value becomes -(c + 5). The equation becomes:

-(c + 5) - 3c = 10 + 2(c - 4) - 6c

Simplifying:

-c - 5 - 3c = 10 - 4c - 8 -4c - 5 = 2 -4c = 7 c = -\dfrac{7}{4}

Case 3: c + 5 < -5

Here, the absolute value becomes -(c + 5). The equation becomes:

-(c + 5) - 3c = 10 + 2(c - 4) - 6c

Simplifying:

-c - 5 - 3c = 10 - 4c - 8 -4c - 5 = 2 -4c = 7 c = -\dfrac{7}{4} (Solution)

We saw that only c = -7/4 satisfies the equation when -5 ≤ c + 5 < 0.

Summary:

Therefore, the only solution to the equation is c = -\dfrac{7}{4}.

This equation involves logarithms in multiple bases. To solve for x, it's generally easier to rewrite everything in terms of a single base.

We can approach this problem by using the following properties of logarithms:

Change of Base Rule: We can change the base of a logarithm using the following rule: logb​(a)=logc​(b)logc​(a)​

Product Rule: The logarithm of a product is the sum of the logarithms of the individual factors: logb​(a⋅b)=logb​(a)+logb​(b)

Let's apply these properties:

Change base of one term: We can rewrite the term log4​(x) using base 3, the same base as the first term:

log4​(x)=log3​(4)log3​(x)​

Substitute and apply product rule: Substitute the rewritten term into the original equation:

log2​(log3​(x))=2⋅log3​(4)log3​(x)​

$$ \log_2(\log_3(x)) = \frac{2 \cdot \log_3(x)}{\log_3(4)}$$

Now we have both logarithms in base 3. However, it's still difficult to solve for x directly.

Here, we can notice something interesting. The left-hand side represents the logarithm of log3​(x) (base 2), while the right-hand side has a term log3​(x) in the numerator. This suggests a potential relationship between x and log3​(x).

Exploring the Relationship:

Let's consider what happens to the value of log3​(x) as x increases:

If x is very small (say, less than 1), then log3​(x) is negative.

As x increases, log3​(x) increases and becomes positive.

As x keeps increasing, log3​(x) also keeps increasing but at a slower rate.

Now, let's think about the logarithm of log3​(x) (base 2).

If log3​(x) is negative, then its logarithm (base 2) is undefined.

As log3​(x) becomes positive and small, its logarithm (base 2) will also be a small positive value (since 2 raised to a small positive power is a bit larger than 1).

As log3​(x) increases further, its logarithm (base 2) will also increase but at a slower rate similar to log3​(x) itself.

Looking at the equation:

The equation suggests that the left-hand side (log of something) needs to be equal to a constant multiple (2) of the right-hand side (something itself). This implies a scenario where the "something" on the right-hand side is a value that, when taking its logarithm (base 2), results in a similar value to itself.

This scenario points us towards a value for x that is very close to, but slightly larger than, 2.

Trying a value:

Let's try plugging in x = 2.5 into the equation:

log3​(2.5)≈0.4 (approximately positive and small)

2⋅log3​(2.5)≈0.8 (approximately positive and small, similar to log3​(2.5))

This supports our guess that x should be close to 2.

Solving for x:

Unfortunately, due to the complexity of logarithms, it's difficult to find an exact solution for x algebraically. However, we can use calculators or computer programs that can handle logarithms with various bases.

Evaluating the original equation with x = 2.5, we get a very close approximation to 0 on both sides (due to the properties of logarithms mentioned earlier). This suggests that x = 2.5 is a good approximation for the solution.

Therefore, the solution for x is approximately x = \boxed{2.5}.

The last two digits of a number depend only on the last two digits of the number itself. Note that the last two digits of the powers of 9 repeat in a cycle of 1: {1,9,81,65,49,21}.

We can start calculating the last two digits of some powers of 9:

91=9 (last two digits: 09)

92=81 (last two digits: 81)

93=729 (last two digits: 29)

94=6561 (last two digits: 61)

We see that the last two digits repeat in a cycle of 4: {09, 81, 29, 61}. Since 4 divides 7, the remainder upon dividing 7 by 4 is the same as the remainder upon dividing 87 by 4.

Because 8 is even, any power of 8 is also even, so 87 has a remainder of 0 when divided by 4. Therefore, the last two digits of 9^(8^7) are the same as the last two digits of 9^4, which is 61​.

We should first note that, because Mark must be on the committee, there are 19 people remaining to choose from. Also, because Mark must be on the committee, so there are 6 spots remaining.

Therefore, the answer is just \(\binom{19}{6}\)

\(=27132\)

\(\frac{128\pi}{3}\)just set 16\sqrt5 equal to the surface area formula and substitute

Catherine rolls a standard 6-sided die six times. If the product of her rolls is 200, then how many different sequences of rolls could there have been? (The order of the rolls matters.)  

The prime factorization of 200 is 2 x 2 x 2 x 5 x 5 but  

that's only five factors and Cat rolled six times so we  

need to add a 1 to the rolls. 

She had to roll these numbers:  1, 2, 2, 2, 5, 5     

How many sequences of these could have occurred.  

The permutation of six things taken six at a time.  

                            ₆P₆ = 720

Oh, she could have rolled 1, 1, 4, 2, 5, 5  etc.      

_.

Ahhh, I see. I didn't think about using the two equations we make and combining them in the end. 

I was more focued on setting a variable, which I see is not needed. 

Thanks CPhill, I understand how to solve this problem now!

Not as bad as you might  think.....

CE = 2sqrt 61  = sqrt 244

AD = sqrt 601 

Note  that we have this system

AD^2  = (BC/2)^2  + (AB)^2

CE^2  = (BC)^2 + (AB/2)^2         simplify

601  = BC^2/4 + AB^2

244 = BC^2 + (AB/2)^2

601 = BC^2 /4  + AB^2

244 = BC^2 + AB^2/4                mutiply the top equation through by -4

-2404 = -BC^2- 4AB^2

244  = BC^2  + AB^2/4     add these

-2160  =  -(15/4)AB^2

(-2160) * (-4/15)  =AB^2  = 576

AB = sqrt (576)  = 24

To find BC

244 = BC^2 + (24/2)^2

244 = BC^2 + 144

100 = BC^2

10= BC

Hypotenuse  =  sqrt [ 24^2 + 10^2 ] =  sqrt [676 ]  =  26

https://web2.0calc.com/questions/help-geometry_154

https://web2.0calc.com/questions/triangle_41241

https://web2.0calc.com/questions/triangle_19840

  A

30

                X

B 90                 C 60

            12

AB =  12sqrt 3

Law of Sines

AB  /  sin BXA  = BX / sin BXA

12sqrt 3 / sin 75  = BX / sin 30

BX  =  6sqrt 3 / sin 75  

[ ABX ]  = (1/2) BX * AB * sin (45) =  

(1/2) (6sqrt 3 / sin 75) ( 12sqrt 3) sin (45)

108 sin (45) / sin(75)  =

216 / (1 + sqrt 3)  ≈   79.06

Think about this...let m  =  1

So   a  =  (1,0)    and b =(0, 1)

These are the same  length  and  are perpendicular

A                        B

                          4

                          P

                          1

D     1    Q  4     C

AB  =  5

AP^2  =  5^2 + 4^2  =  41

AD  = 5

AQ^2  =  5^2 + 1^2  =  26

PQ^2  = 4^2 + 1^2  = 17

Law  of Cosines

PQ^2  = AP^2 + AQ^2  -  2 (AP* AQ) cos (PAQ)

[PQ^2 - AP^2 - AQ^2 ] / [ -2 sqrt (AP * AQ) ]  = cos PAQ

[ 17  - 41 - 26 ] / [ -2 sqrt (26 * 41) ]  = cos PAQ  = 25 / sqrt 1066 =  cos PAQ

sin PAQ  =  sqrt [ 1066 - 25^2 ] / sqrt 1066  = sqrt (441) / sqrt 1066  =   21  / sqrt 1066

How did you get a  = ( 0, m) and b = (m , 0). Why the 0?

The triangle is equilateral...AD is a bisector and  the altitude of ABC

                      A

                     24

B                   D                 C

Angle  BAD  = 30

BD  =  24/sqrt 3

BC  =  2 * 24 /sqrt 3 =   48/ sqrt 3 =   16sqrt 3

[ABC ]  = (1/2) BC * AD  =  (1/2) (16sqrt 3) (24)  =  192 sqrt 3