All Answers 358087 Answers

Here are the steps to the solution:

1)Draw the triangle.

2) Mark in all the angles. You have enough information to find the whole lot just using the angle sum of a triangle.

3) Now just by looking at what kind of triangles you have and you can see the length of AC

4) Use sin rule to find BC    [AC would be ok too but it is easier to get an exact area using BC]

5) Now you have all the angles and 2 sides so use the formula to find the area of ABC.    ( area= 0.5xysinZ,  if it had beeen triangle XYZ)

My answer is different from all the others but I could easily have made a careless error.

If you are really interested in learning then follow my instructions and see where it takes you.  

You can ask questions along the way if you want to.

Sorry, that's also incorrect! Thanks for trying tho! I still need helpppp

I didn't

the question is incomplete.

Ask one question and learn from the answer.

Guest one is right.

We are not here to do your homework for you.

(b) What is the probability that Valeria has three heads on her first turn?
Ans:- Assuming that Valeria is flipping a fair coin, the probability of getting a heads on any given flip is 0.5 (or 1/2).

The probability of getting three heads in a row on the first three flips is the product of the probabilities of each individual flip coming up heads, since the flips are independent of each other:

P(three heads on first turn) = P(heads) * P(heads) * P(heads) = 0.5 * 0.5 * 0.5 = 0.125 or 12.5%

Therefore, the probability that Valeria has three heads on her first turn is 0.125 or 12.5%.

This list is that of the natural counting numbers, which is free of squares, cubes and perfect powers:

(2, 3, 5, 6, 7, 10, 11, 12, 13, 14, 15,  17, 18, 19, 20, 21, 22, 23, 24, 26,  28, 29, 30, 31, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113)==100 terms.

In this list, the 100th term is 113.

This list of counting numbers, whose divisors or factors, are square-free, cube-free and perfect power-free.

(2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 46, 47, 51, 53, 55, 57, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97, 101, 102, 103, 105, 106, 107, 109, 110, 111, 113, 114, 115, 118, 119, 121, 122, 123, 127, 129, 130, 131, 133, 134, 137, 138, 139, 141, 142, 143, 145, 146, 149, 151, 155, 157, 159, 161, 163, 165, 166)==100 terms.

In this list, the 100th term is 166.

Bruh this is easy the answer is the answer of A/B= 87.34433334578976 you have to do the math in a calculator your welcome hope u get a good grade buddy

10 - 99 : 11
100 - 999 : 112
1000 - 9999 : 1124
10000 - 99999 : 11248
100,000 - 999,999 : 112,496
1000000 - 9999999 : 1124992
10000000 - 99999999 : 11249985
100000000 - 999999999 : 112499976
1000000000 - 9999999999 : 1124999972

A baking club wants to form an executive committee.  There are 10 people in the baking club, including Mark.  In how many ways can the baking club form an executive committee with 3 people?

To form an executive committee of 3 people out of 10 members, we need to select 3 people out of 10, and the order in which they are selected does not matter. This can be calculated using the combination formula:

makefileCopy code

nCk = n! / (k! * (n-k)!)

where n is the total number of people (10), and k is the number of people we want to choose (3).

Plugging in the values, we get:

markdownCopy code

10C3 = 10! / (3! * (10-3)!) = 120

Therefore, there are 120 possible committees that can be formed.

sorry, this is also wrong

We need to find the number of values of x that satisfy the equation f(f(x)) = 5. Let's start by solving this equation.

Since f(x) is defined piecewise, we need to consider two cases:

Case 1: x < -4 In this case, f(x) = x + 3. Therefore, f(f(x)) = f(x + 3) = (x + 3) - 3 = x - 6. So we need to solve the equation f(f(x)) = x - 6 = 5, which gives x = 11.

Case 2: x >= -4 In this case, f(x) = x^2 - 4. Therefore, f(f(x)) = f(x^2 - 4) = (x^2 - 4)^2 - 4. So we need to solve the equation f(f(x)) = (x^2 - 4)^2 - 4 = 5. Expanding the left side, we get (x^4 - 8x^2 + 12) = 0. This is a quartic equation that can be factored as (x^2 - 2)(x^2 - 6) = 0. Therefore, the solutions are x = sqrt(2), -sqrt(2), sqrt(6), and -sqrt(6).

We need to check which of these solutions also satisfy the condition x >= -4 since this is the second case of the piecewise definition of f(x).

We have:

sqrt(6) < -2 < -4, so -sqrt(6) is not a valid solution. sqrt(6) > 2 > -4, so sqrt(6) is a valid solution.

Therefore, there are two values of x that satisfy the equation f(f(x)) = 5: 11 and sqrt(6).

12,  21,  111 ==3 such integers !

Area ==a^2 * sin(θ)

Area ==10^2 * sin(120)

Area ==100 x  sqrt(3) / 2

Area ==50 sqrt(3) 

We are not here to do your homework for you.

The initial mixture is 60% water and 40% lemon juice. This means that the jug contains 60/100 * 100 ounces = 60 ounces of water, and 40/100 * 100 ounces = 40 ounces of lemon juice.

When 60% of the lemonade is poured out, the jug now contains 40% of its original volume. Therefore, the jug contains

40/100 * 100 ounces = 40 ounces

of liquid, of which 60% is water and 40% is lemon juice. This means that the jug contains:

60/100 * 40 ounces = 24 ounces of water 40/100 * 40 ounces = 16 ounces of lemon juice

Now 8 ounces of lemon juice are added to the jug, bringing the total amount of lemon juice to:

16 ounces + 8 ounces = 24 ounces

The total volume of the mixture is now:

24 ounces of lemon juice + 24 ounces of water = 48 ounces

Therefore, the percentage of the mixture that is now lemon juice is:

(24 ounces / 48 ounces) * 100% = 50%

So, the mixture now contains 50% lemon juice.

The side length of the square is 3*sqrt(2).

Let $n$ be a positive integer satisfying the given conditions, and let $a$ be the number of 1's in the base-2 representation of $n.$ Then $n$ has $3-a$ 2's in its base-2 representation. Since $n$ has a sum of digits of 3, we know that $a + (3-a)\cdot 2 = 3,$ or $a = 3.$ Thus, $n$ must have three 1's and no 2's in its base-2 representation.

Conversely, any positive integer with three 1's and no 2's in its base-2 representation is a solution to the given conditions. There are $\binom{3}{0} = 1$ ways to place zero 2's among three 1's, so the number of positive integers that satisfy the conditions is $\boxed{1}.$

The side length of the square is 1 + 2*sqrt(3).

[asy]
size(4cm);
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
draw((0,0)--(0,-1)--(1,-3^.5)--(2,-1)--(2,0));
draw((2,0)--(3,1)--(2,2)--(1,3)--(0,2));
draw((0,2)--(-1,1)--(-3^.5,0)--(-1,-1)--(0,-2));
draw((0,-2)--(1,-3)--(2,-2)--(3,-1)--(2,0),dashed);

[/asy]

Let $O$ be the center of the square. Since the side length of the square is $s$, the distance from $O$ to any vertex of the square is $\sqrt{s^2 + s^2} = s\sqrt{2}.$

Let $A,$ $B,$ $C,$ $D$ be the vertices of the square, in that order. Let $E,$ $F,$ $G,$ $H$ be the centers of the equilateral triangles with $\overline{AB},$ $\overline{BC},$ $\overline{CD},$ $\overline{DA}$ as bases, respectively.

Let $X$ be the intersection of $\overline{EF}$ and $\overline{OG},$ and let $Y$ be the intersection of $\overline{GH}$ and $\overline{OF}.$

[asy]
size(6cm);
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
draw((0,0)--(0,-1)--(1,-3^.5)--(2,-1)--(2,0));
draw((2,0)--(3,1)--(2,2)--(1,3)--(0,2));
draw((0,2)--(-1,1)--(-3^.5,0)--(-1,-1)--(0,-2));
draw((0,-2)--(1,-3)--(2,-2)--(3,-1)--(2,0),dashed);
draw((0,0)--(1,-3^.5)--(2,0),dashed);
draw((0,2)--(-1,1)--(-3^.5,0)--(0,-1)--(0,0),dashed);
draw((2,0)--(3,1)--(2,2)--(1,1)--(1,0),dashed);
label("$A$",(0,2),NW);
label("$B$",(2,2),NE);
label("$C$",(2,0),SE);
label("$D$",(0,0),SW);
label("$E$",(1,-3^.5),S);
label("$F$",(2,0),E);
label("$G$",(-1,1),W);
label("$H$",(0,2),N);
label("$O$",(1,1),NW);
label("$X$",(5/3,-1/3),S);
label("$Y$",(4/3,4/3),E);
draw((1,1)--(5/3,-1/3),dashed);
draw((1,1)--(4/3,4/3),dashed);
[/asy]

Since $\triangle ABE$ is equilateral, $AE = BE = s.$ Since $O$ is the center of the square, $OE = OD - DE = s - \frac{s}{2} = \frac{s}{2}.$ Therefore, $OX = OE + EX = \frac{s}{2} + \frac{s\sqrt{3}}{2} = \frac{s(1+\sqrt{3})}{2}.$ Similarly, $OY = OE + EY = \frac{s}{2} + \frac{s\sqrt{3}}{2} = \frac{s(1+\sqrt{3})}{2}.$

Then, the side length of the large square is $XY = OX + OY = s(1 + \sqrt{3}).$

Therefore, the answer is $\boxed{s(1 + \sqrt{3})}.$

[asy]
size(4cm);
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
draw((0,0)--(0,-1)--(1,-3^.5)--(2,-1)--(2,0));
draw((2,0)--(3,1)--(2,2)--(1,3)--(0,2));
draw((0,2)--(-1,1)--(-3^.5,0)--(-1,-1)--(0,-2));
draw((0,-2)--(1,-3)--(2,-2)--(3,-1)--(2,0),dashed);

[/asy]

Let $O$ be the center of the square. Since the side length of the square is 2, the distance from $O$ to any vertex of the square is $\sqrt{2^2 + 2^2} = 2\sqrt{2}.$

Let $A,$ $B,$ $C,$ $D$ be the vertices of the square, in that order. Let $E,$ $F,$ $G,$ $H$ be the centers of the equilateral triangles with $\overline{AB},$ $\overline{BC},$ $\overline{CD},$ $\overline{DA}$ as bases, respectively.

Let $X$ be the intersection of $\overline{EF}$ and $\overline{OG},$ and let $Y$ be the intersection of $\overline{GH}$ and $\overline{OF}.$

[asy]
size(6cm);
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
draw((0,0)--(0,-1)--(1,-3^.5)--(2,-1)--(2,0));
draw((2,0)--(3,1)--(2,2)--(1,3)--(0,2));
draw((0,2)--(-1,1)--(-3^.5,0)--(-1,-1)--(0,-2));
draw((0,-2)--(1,-3)--(2,-2)--(3,-1)--(2,0),dashed);
draw((0,0)--(1,-3^.5)--(2,0),dashed);
draw((0,2)--(-1,1)--(-3^.5,0)--(0,-1)--(0,0),dashed);
draw((2,0)--(3,1)--(2,2)--(1,1)--(1,0),dashed);
label("$A$",(0,2),NW);
label("$B$",(2,2),NE);
label("$C$",(2,0),SE);
label("$D$",(0,0),SW);
label("$E$",(1,-3^.5),S);
label("$F$",(2,0),E);
label("$G$",(-1,1),W);
label("$H$",(0,2),N);
label("$O$",(1,1),NW);
label("$X$",(5/3,-1/3),S);
label("$Y$",(4/3,4/3),E);
draw((1,1)--(5/3,-1/3),dashed);
draw((1,1)--(4/3,4/3),dashed);
[/asy]

Since $\triangle ABE$ is equilateral, $AE = BE = 2.$ Since $O$ is the center of the square, $OE = OD - DE = 2 - 1 = 1.$ Therefore, $OX = OE + EX = 1 + \frac{2}{\sqrt{3}}.$ Similarly, $OY = OE + EY = 1 + \frac{2}{\sqrt{3}}.$

Then, the side length of the large square is $XY = OX + OY = 2 + \frac{4}{\sqrt{3}}.$

Therefore, the answer is $\boxed{2 + \frac{4}{\sqrt{3}}}.$

We can treat the block of 5 scientists as a single entity and arrange them in a line first. This can be done in 5! ways. Then, we can arrange the other 12 people around the circular table in (12-2+1)! = 11! ways (since we need to leave 2 seats for the mathematicians who will sit together).

Finally, since the table is circular, we need to divide by 14 (the number of positions we can rotate the table to get the same arrangement) to get the total number of distinct arrangements:

5! * 11! / 14

Simplifying, we get:

(5 * 4 * 3 * 2 * 1) * (11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / 14

= 120 * 39,916,800 / 14

= 342,720,000

Therefore, there are 342,720,000 different arrangements possible if the scientists must all sit together.

We can solve this system of equations using Gaussian elimination. First, we can add 4 times the first equation to the second equation to eliminate the x term:

x + y + 3z = 10
(4)(x + y + 3z) + (-4x + 8y + 5z) = (4)(10) + 7
kx + z = 3

Simplifying, we get:

x + y + 3z = 10
20y + 17z = 47
kx + z = 3

Next, we can subtract 20 times the third equation from the second equation to eliminate the y term:

x + y + 3z = 10
-20kx - 20z + 17z = 47 - 60k
kx + z = 3

Simplifying, we get:

x + y + 3z = 10
(k - 3)z = 60k - 47
kx + z = 3

If this system has no solutions, then the equations must be inconsistent, which means that there is no value of k that satisfies all three equations. 

In particular, the second equation cannot be satisfied for any value of k, since the left-hand side is always a multiple of 17, while the right-hand side is not a multiple of 17.

Therefore, we can set the coefficients of z equal to each other and solve for k:

k - 3 = 60k - 47

Simplifying, we get:

59k = 50

k = 50/59

So the value of k that makes the system inconsistent is k = 50/59.

Notice that if we have some sequence of the numbers 1, 2, 3, 4, 5, 6, and 7, then the parity (evenness or oddness) of every odd-indexed term (the 1st, 3rd, 5th, and 7th terms) must be the same, and the parity of every even-indexed term (the 2nd, 4th, and 6th terms) must also be the same.

Let's assume without loss of generality that the 1st term is odd. Then, the 2nd term must be even, the 3rd term must be odd, the 4th term must be even, and so on.

There are 4 odd numbers and 3 even numbers, so there are 4 choices for the 1st term, and once the 1st term is chosen, there are 3 choices for the 2nd term (it must be an even number not equal to the 1st term), 3 choices for the 3rd term (it must be an odd number not equal to the 2nd term), and so on.

Therefore, the total number of arrangements is:

4 * 3 * 3 * 2 * 2 * 1 * 1 = 144

So there are 144 arrangements of the numbers 1, 2, 3, 4, 5, 6, and 7 where any sum of any two adjacent numbers is odd.