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ElenaW [278]
2 years ago
9

Plzzzzz helpppp!!!!!!

Mathematics
2 answers:
disa [49]2 years ago
8 0

Answer:

x=11

Step-by-step explanation:

because a each angle of a square or rectangle is 90° which means you add 9 to 90 and then divided 99 by 9 which is 11

oksano4ka [1.4K]2 years ago
7 0

Answer:

x = 11

Step-by-step explanation:

9x - 9 = 90

    + 9  + 9

9x = 99

9x/9  99/9

x = 11

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Find the area of the following shape, given its curves are made from parts of circles with diameters indicated on the diagram.
Alexxandr [17]

Answer:

222.51 (when rounded)

Step-by-step explanation:

Pi x r squared is the formula to find the area of a circle

153.94 + 12.57 + 56

= 222.51

You basically find the area of each circle by find their diameters. So half of the length shown on each circle.

7 0
2 years ago
What is W in this equation <br> -w–11≤w–5
zhannawk [14.2K]

Step-by-step explanation:

-w - 11 < or = w - 5

-2w - 11 < or = -5

-2w < or = -16

w > or = -8

5 0
3 years ago
SOMEONE HELP PLEASE!
WINSTONCH [101]

Answer:

x=48 and y is 132 my guy u welcome

6 0
3 years ago
Read 2 more answers
Figure A is a scale image of Figure B.<br><br> What is the value of x?<br> PS. it's not 17
Oksanka [162]

Step-by-step explanation:

24÷(45÷25)= 21

x=21

Hope this helped!

4 0
2 years ago
Read 2 more answers
Find the remainder when 1 + 2 + 2^2 + 2^3 + ... + 2^100 is divided by 7.<br><br> Thanks in advance!
Firlakuza [10]

Answer:

3.

Step-by-step explanation:

This is a geometric series so the sum is:

a1 * r^n - 1 / (r - 1)

= 1 * (2^101 -1) / (2-1)

= 2^101 - 1.

Find the remainder when 2^101 is divided by 7:

Note that 101 = 14*7 + 3 so

2^101 = 2^(7*14 + 3) =  2^3  * (2^14)^7 = 8 * (2^14)^7.

By Fermat's Little Theorem  (2^14) ^ 7 = 2^14 mod 7 = 4^7 mod 7.

So 2^101 mod 7 = (8 * 4^7) mod 7

                           = (8 * 4) mod 7

                            = 32 mod 7

                            = 4 = the remainder when 2^101 is divided by 7.

So the remainder when 2^101- 1 is divided by 7 is 4 - 1 = 3..

4 0
2 years ago
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