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2 years ago

The points (-8, -4) and (0, 2) are endpoints of a diameter of a circle. Identify the standard equation of the circle.

Mathematics

1 answer:

Mariana [72]2 years ago

3 0

Answer:

the first one i think im not 100% sure havent worked on this much

Step-by-step explanation:

You might be interested in

Anybody know the answer?

leonid [27]

Answer:

The answer to your question is AD = 74

Step-by-step explanation:

Data

AB = CD = 10

BC = DE = 54

DE = 54

CD = 10

Process

From the picture we know that

AD = AB + BC + CD

Substitution

AD = 10 + 54 + 10

Simplification and result

AD = 74

7 0

3 years ago

Why is a number raised to the zero power ALWAYS one and never any other number?

alexdok [17]

Do you know how to simplify, let's say for example, x⁵/x² ?

When the bases are the same, you can just subtract the exponent in
the denominator from the exponent in the numerator. So x⁵/x² = x³ .

Now look at x⁷/x⁷ . From that same handy tip, x⁷/x⁷ = x⁰ .

BUT ... any fraction with the same number on top and bottom
is equal to ' 1 '. So x⁰ = 1 , no matter what 'x' happens to be.

Does that do anything for you ?

5 0

3 years ago

Which equation represents the function notation for the equation y = 5x + 4?

alexandr402 [8]

Answer:

f(x) = 5x + 4

Step-by-step explanation:

5 0

2 years ago

A large can of soup contains 19 fl oz A serving is about eight Oz How many Canz should you buy if you want to serve How many Can

stiv31 [10]

Answer:

2.95 cans of soup

Step-by-step explanation:

A large can of soup contains 19 fl oz A serving is about eight Oz How many Canz should you buy if you want to serve How many Cans should you buy if you want to serve 7 people

Step 1

From the above question:

1 serving = 1 Person

1 person = 8oz

7 persons = x

Cross Multiply

x = 7 × 8oz

x = 56 oz

Step 2

56 oz will serve 7 people

From the question also, we are told that:

19 oz = 1 can of soup

56 oz = x

Cross Multiply

19x = 56 × 1

x = 56 /19

x =2.9473684211 cans of soup

Approximately

2.95 cans of soup will serve 7 peop

5 0

3 years ago

The nine squares of a 3-by-3 chessboard are to be colored red and blue. The chessboard is free to rotate but cannot be flipped o

nignag [31]

Answer:

$a_n = 2^{\frac{n^2-1}{4} + 1} + \frac{2^{n^2} - \, 2^{\frac{n^2-1}{4} + 1}}{4}$

For n = 3, there are 134 possibilities

Step-by-step explanation:

First, lets calculate the generating function.

For each square we have 2 possibilities: red and blue. The Possibilities between n² squares multiply one with each other, giving you a total of 2^n² possibilities to fill the chessboard with the colors blue or red.

However, rotations are to be considered, then we should divide the result by 4, because there are 4 ways to flip the chessboard (including not moving it), that means that each configuration is equivalent to three other ones, so we are counting each configuration 4 times, with the exception of configurations that doesnt change with rotations.

A chessboard that doesnt change with rotations should have, in each position different from the center, the same colors than the other three positions it could be rotated into. As a result, we can define a <em>symmetric by rotations chessboard</em> with only (n²-1)/4 + 1 squares (the quarter part of the total of squares excluding the center plus the center).

We cocnlude that the total of configurations of symmetrical boards is $2^{\frac{n^2-1}{4} + 1}$

Since we have to divide by 4 the rest of configurations (because we are counted 4 times each one considering rotations), then the total number of configutations is

$a_n = 2^{\frac{n^2-1}{4} + 1} + \frac{2^{n^2} - \, 2^{\frac{n^2-1}{4} + 1}}{4}$

If n = 3, then the total amount of possibilities are

$a_3 = 2^{\frac{3^2-1}{4} + 1} + \frac{2^{3^2} - \, 2^{\frac{3^2-1}{4} + 1}}{4} = 134$

3 0

3 years ago