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algol13
3 years ago
8

Find the indicated side of the right triangle.

Mathematics
1 answer:
Advocard [28]3 years ago
8 0

Answer:

?=9

Step-by-step explanation:

We have a special right triangle, the 45-45-90 triangle.

In a 45-45-90 triangle, the side lengths are x, and the hypotenuse is x√2.

Since we know that the side length is 9, this means that:

x=9\\\text{Hypotenuse}=x\sqrt2\\\text{Hypotenuse}=9\sqrt2

The ? is 9.

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What is the range of y=5(11)x
dem82 [27]

Answer:

Domain: ( − ∞ , ∞ ) ,  { x | x ∈ R}

Range: ( − ∞ , ∞ ) ,  { y | y ∈ R}

Step-by-step explanation:

We can simplify to understand how to answer the question better.

<em>5 * 11x = 55x</em>

<u>Function range definition:</u><u> </u>

The set of values of the dependent variable for which a function is defined.

  • The range of polynomials with odd degree is all real numbers.

<u>So the solution is.. </u>

-∞<x< ∞

<u>And the interval notation is..</u>

(- ∞. ∞)

7 0
3 years ago
Question
Gnom [1K]

Answer:

ask to your teachers and you will be finish the work

3 0
2 years ago
Some people think it is unlucky if the 13th day of a month falls on a Friday. Show that in every calendar year (non leap or leap
MakcuM [25]
We will set a variable, d,  to represent the day of the week that January starts on.  For instance, if it started on Monday, d + 1 would be Tuesday, d + 2 would be Wednesday, etc. up to d + 6 to represent the last day of the week (in our example, Sunday).  The next week would start over at d, and the month would continue. For non-leap years:
If January starts on <u>d</u>, February will start 31 days later.  Following our pattern above, this will put it at <u>d</u><u> + 3</u> (28 days would be back at d; 29 would be d+1, 30 would be d+2, and 31 is at d+3).  In a non-leap year, February has 28 days, so March will start at <u>d</u><u>+3</u> also.  April will start 31 days after that, so that puts us at d+3+3=<u>d</u><u>+6</u>.  May starts 30 days after that, so d+6+2=d+8.  However, since we only have 7 days in the week, this is actually back to <u>d</u><u>+1</u>.  June starts 31 days after that, so d+1+3=<u>d</u><u>+4</u>.  July starts 30 days after that, so d+4+2=<u>d</u><u>+6</u>.  August starts 31 days after that, so d+6+3=d+9, but again, we only have 7 days in our week, so this is <u>d</u><u>+2</u>.  September starts 31 days after that, so d+2+3=<u>d</u><u>+5</u>.  October starts 30 days after that, so d+5+2=d+7, which is just <u>d</u><u />.  November starts 31 days after that, so <u>d</u><u>+3</u>.  December starts 30 days after that, so <u>d</u><u>+5</u>.  Remember that each one of these expressions represents a day of the week.  Going back through the list (in numerical order, and listing duplicates), we have <u>d</u><u>,</u> <u>d,</u><u /> <u>d</u><u>+1</u>, <u>d</u><u>+2</u>, <u>d+3</u><u>,</u> <u>d</u><u>+3</u>, <u>d</u><u>+3</u>, <u>d</u><u>+4</u>, <u>d</u><u>+5</u>, <u>d</u><u>+5</u>, <u /><u /><u>d</u><u>+6</u><u /><u /> and <u>d</u><u>+6</u>.  This means we have every day of the week covered, therefore there is a Friday the 13th at least once a year (if every day of the week can begin a month, then every day of the week can happy for any number in the month).  
For leap years, every month after February would change, so we have (in the order of the months) <u></u><u>d</u>, <u>d</u><u>+3</u>, <u>d</u><u>+4</u>, <u>d</u><u />, <u>d</u><u>+2</u>, <u>d</u><u /><u>+5</u>, <u>d</u><u />, <u>d</u><u>+3</u>, <u>d</u><u /><u>+6</u>, <u>d</u><u>+1</u>, <u>d</u><u>+4</u>, a<u />nd <u>d</u><u>+</u><u /><u /><u>6</u>.  We still have every day of the week represented, so there is a Friday the 13th at least once.  Additionally, none of the days of the week appear more than 3 times, so there is never a year with more than 3 Friday the 13ths.<u />
5 0
3 years ago
Prove that A is a subset of B if and only if B complement is a subset of A complement.
WITCHER [35]
Please read the attached file

3 0
3 years ago
13. In the triangle below, the sides have length x + 3, 3x - 1. and 4x. Determine the range of possible
11Alexandr11 [23.1K]

Answer:

Hence as long as x > 2/3, we can form a triangle from the three given sides.

Step-by-step explanation:

Given:

Length of 3 sides of triangle are x+3, 4x,3x-1

Solution:

From the length of the 2nd side 4x, we know that x > 0

Let this be 1 st statement.

Now from the triangle inequality we can say that;

x + 3 + 4x > 3x -1\\5x+3>3x-1\\5x-3x>-1-3\\2x>-2\\x>-1

No new information from this because of the 1st statement above.

Also,  

4x + 3x - 1 > x +3\\7x-1>x+3\\7x-x>3+1\\6x>4\\x>\frac{4}{6}\\\\x>\frac{2}{3}

Lastly,

x+3 +3x - 1 > 4x\\4x-2>4x\\4x-4x>2\\0>2

and again no new information is obtained from this inequality.

Hence as long as x > 2/3, we can form a triangle from the three given sides.

7 0
3 years ago
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