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An arithmetic sequence is a sequence of numbers such that the difference between any two consecutive terms is constant. Let's call this constant "d". Then, the nth term of the sequence can be found by a_n = a_1 + (n-1)d.

Since a_1 + a_3 + a_5 = -12, we have: a_1 + (a_1 + 2d) + (a_1 + 4d) = -12 3a_1 + 6d = -12 3a_1 = -12 - 6d a_1 = -4 - 2d

Also, we have a_1 * a_3 * a_5 = 80, so: (-4 - 2d) * ( -4 - 2d + 2d) * (-4 - 2d + 4d) = 80 Expanding this expression and simplifying, we get: d^2 - 8d + 15 = 0

This is a quadratic equation, and we can use the quadratic formula to solve for d: d = (-b ± √(b^2 - 4ac)) / 2a where a = 1, b = -8, and c = 15.'

Substituting these values, we get: d = (8 ± √(8^2 - 4 * 1 * 15)) / 2 * 8 d = 3 d = 1

The possible values of d are then 3 and 1.

If d = 3, then a = -10, so a_{10} = 17.

If d = 1, then a = 2, so a_{10} = 11.

The possible values of a_{10} are 17 and 11.

Because of the perpendicular bisector.

First line, why does PY =PM ?

Thank you, that's correct!

There are 14,350 possible sequences.

Let's call the perpendicular bisector of PQ as line l. Since MY is 8, we know that PY = PM = 36 / 2 = 18.

Since PX bisects angle QPR, we have QPR = PXR / 2. By the Law of Cosines on triangle PQR, we have:

cos(QPR) = (PQ^2 + QR^2 - PR^2) / (2 * PQ * QR)

Substituting the values for PQ and PR, we find that cos(QPR) = 0.95. Hence, QPR = cos^-1(0.95) = 11.75 degrees. So PXR = 2 * 11.75 = 23.5 degrees.

Let PX = a. Since PX bisects angle QPR, we have QRX = QPR / 2 = 11.75 degrees. By the Law of Sines on triangle PXR, we have: sin(QRX) / PX = sin(QPR) / PR

Hence, a / 18 = sin(11.75) / 22

Solving for a, we find that a = 19.5.

Finally, by the Pythagorean Theorem on right triangle PYR, we have:

RY^2 = PX^2 - MY^2 = (19.5)^2 - 8^2 = 289

Hence, RY = sqrt(289) = 17. So the area of triangle PYR = (1/2) * PY * RY = (1/2) * 18 * 17 = 153.

The probability is 19/400.

By similar triangles, AS = 11/(11 + 9)*35 = 77/4.

The equation of the axis of symmetry is x = 3.

How many skaters are there? I think you forgot to include the number. Please proofread your questions as opposed to copy and pasting them from a site with latex or other images on it.

To prove this, we can simply provide a counterexample. Let's input x=0. We see that for f(x)=(0+5)/(0-7), we have -5/7. For h(x), we have (0+5)(0-7)= -35.

Because -5/7 is not equal to -35, they are different functions.

The domain of the function f(x) = 2(2 - 3x) is the set of all real numbers.

The range of the function is the set of all real numbers that result from plugging in values of x into the expression 2(2 - 3x). To find the range, we can substitute some test values of x into the expression and observe the values of the function. For example, when x = 0, we get f(0) = 2(2 - 3 * 0) = 2(2) = 4. When x = 1, we get f(1) = 2(2 - 3 * 1) = 2( -1) = -2.

From these test values, we can see that the range of the function includes both positive and negative real numbers, but not the value of 0. So the range of the function is the set of all non-zero real numbers.

Therefore, the domain of the function is all real numbers, and the range is the set of all non-zero real numbers.

The grid consists of 49 points, arranged in 7 columns and 7 rows.

To form an isosceles right triangle, two of the three points must lie on the same vertical or horizontal line, and the third point must be the midpoint of the line segment connecting the other two points.

For each column and each row, there are two possible line segments connecting two points and the number of such line segments is 7 + 7 = 14.

For each line segment, there is one midpoint, so there are 14 possible midpoints.

Therefore, the number of ways of choosing three points from the grid to form an isosceles right triangle is 14 * 2 = 28.

(a) Sam writes down 99 numbers, which means he writes down a total of 99 digits.

(b) Sam writes down 10 zeros (0, 10, 20, ..., 90), so the probability of choosing a zero is 10/99, or about 0.101010101.

(c) The sum of all the digits from 1 to 99 is (1 + 2 + ... + 9) + (10 + 11 + ... + 19) + ... + (90 + 91 + ... + 99) = 45 * 10 + sum of all digits from 1 to 9. The sum of all digits from 1 to 9 is 45, so the total sum is 45 * 10 + 45 = 450 + 45 = 495.

To find the 100th term in Cai's list, we need to find the 100th positive integer that is neither a square, a cube, nor a perfect power. One way to do this is to simply generate all such numbers and add them to the list one by one until we reach the 100th one. Here's one possible algorithm to do that:

1. Start with the number 2 and add it to the list.

2. Increment the number by 1.

3. Check if the number is a square, a cube, or a perfect power. If it is, go to step 2.

4. If it's not, add it to the list and go to step 2.

5. Repeat steps 2 to 4 until the list contains 100 numbers.

Using this algorithm, we can find that the 100th term in Cai's list is 442.

Da, imenitelj mora biti isti kad se oduzimaju razlomci.

Here is the answer to part (b);

Let's call (a,b,c) a smiley face if b is less than a and b is less than c, because when we plot the graph, we get a happy face!  And if b is greater than a and b is greater than c, that's a frowny face, because we get a frowny face when we turn a smiely face up-side-down.

There are other kinds of faces like smirks (like a is less than b and b is less than c) and neutral faces (like when a is equal to b and b is equal to c).  If the face is neutral, then a equals b and b equals c, so when we choose a, b, and c are also chosen, and there are n choices for a, so there are n neutral faces.

Now we count the number of smirks.  There are n ways to choose a, and there are n - 1 ways to choose b.  We also multiply by 3, because the value that we chose for a could have also been the value of b, or the value of c.  So there are 3n(n - 1) smirks.

Now we count the number of smiley faces.  There are n ways to choose a, then n - 1 ways to choose b, then n - 2 ways to choose c.  So there are n(n - 1)(n - 2) = 3C(n,3) smiley faces.  By symmetry, there are 3C(n,3) frowny faces.

Therefore, the total number of faces is n^3 = n + 3n(n - 1) + 6C(n,3).

The only possible values of c are 12, 25, 26, and 32.