Please Help its past due!

Mathematics

2 answers:

andreev551 [17]3 years ago

6 0

HEY THERE !!

YOUR ANSWER IS IN ATTACHMENT !!

See both attachments.

salantis [7]3 years ago

3 0

Draw or Cut two similar squares with sides $b+c$ units long.

Draw or cut four pairs of similar right triangles with side lengths as indicated in the diagram.

Now arrange the similar triangles at the corners of the squares such that the sides $b$ of one similar triangle plus the side $c$ of a second similar triangle coincides with the length of the square.

We do another arrangement of the similar triangles. This time arrange another 4 similar triangles in the opposite corners, such that each pair forms a square.

Now comparing the two different arrangements we got two different areas that are equal.

The area of the uncovered squares in the first arrangement is $a^2$

The area of the two uncovered squares in the second arrangement is $b^2+c^2$

Equating the two areas gives the Pythagoras Theorem

$a^2=b^2+c^2$

Note that $a$ is the hypotenuse, $b$ and $c$ are two shorter sides of the similar right triangles.

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Find five consecutive integers such that: "the sum of the first and 4 times the third is equal to 60 less than 3 times the sum o

Monica [59]

Consecutive integers are integers that come one after another in order. For example, 2, 3, and 4 are all consecutive integers, as well as 100, 101, and 102. In general, if we have some integer n, the consecutive integers following it will be n + 1, n + 2, n + 3, etc. Here, we have five. If we

Call the first integer n,
Then the second integer will be n + 1,
The third will be n + 2,
The fourth will be n + 3,
And the fifth will be n + 4.

Now, let's break down the equation in English. We'll use #1, #2, #3, #4, and #5 as stand-ins for the five integers just so we can get things set up, then we'll replace them with the expressions we came up with in the first part. Starting from the beginning of the statement, let's translate words to symbols:

"The sum of the first and 4 times the third" → #1 + 4 × #3
"is equal to" → = (maybe obvious, but still!)
"60 less than 3 times the sum of the second, fourth, and fifth" → 3 × (#2 + #4 + #5) - 60

Putting that all together, we get the equation

#1 + 4 × #3 = 3 × (#2 + #4 + #5) - 60.

Replacing our stand-ins for the expressions we came up with, it becomes

n + 4(n + 2) = 3[(n + 1) + (n + 3) + (n + 4)] - 60

We can now start to simplify our equation to find n (the first integer) and the rest will follow easily from that. I'll start by focusing on the left side. We can distribute and combine like terms like so:

n + 4(n+2) = n + 4n + 8 = 5n + 8

Dealing with the right, we can get rid of some redundant parentheses and carry out the same process:

3[(n + 1) + (n + 3) + (n + 4)] - 60 = 3(n + 1 + n + 3 + n + 4) - 60

= 3(3n + 8) - 60 = 9n + 24 - 60 = 9n - 36

Our simplified equation now becomes

5n + 8 = 9n - 36.

Subtracting 5n from either side:

8 = 4n - 36

Adding 36 to either side:

44 = 4n

And dividing either side by 4:

11 = n.