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lora16 [44]
2 years ago
12

30% of___ is 15 ?????

Mathematics
1 answer:
tensa zangetsu [6.8K]2 years ago
7 0

Answer:

15 is 30% of 50

Step-by-step explanation:

We have, 30% × x = 15

Multiplying both sides by 100 and dividing both sides by 30,

we have x = 15 × 100/30

x = 50

If you are using a calculator, simply enter 15×100÷30, which will give you the answer.

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2 years ago
Find the AREA and PERIMETER of the triangle
iVinArrow [24]

Answer:

Area=99

Step-by-step explanation:

width×Length

9×9+13

=198

198÷2=99

8 0
2 years ago
Vertex Z of trapezoid WXYZ has coordinates (−9, 9).
Taya2010 [7]
The coordinates would surely be -9,15

3 0
3 years ago
Solve using square root method
avanturin [10]
\frac{1}{2}(x+5)^2 + 9 = -15

\frac{(x+5)^2+18}{2} =  \frac{-30}{2}

Cancel out the denominators.

(x+5)² + 18 = -30

(x+5)² = -30 - 18

(x+5)² = -48

The equation is impossible, we have a negative number.

There isn't a number that squared gives a negative number (the square is always positive).

√(x+5)² = +/- √-48

x+5 = +/- √-48

IMPOSSIBLE

You can write the result in imaginary numbers

x+5=+/- 4i√3
6 0
2 years ago
Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-ma
pickupchik [31]

Answer:

Answer explained below

Step-by-step explanation:

Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-mails containing the same virus, after which the virus disables itself on that machine. (1) Write a recursive definition (i.e. recurrence relation) to show how many spam emails will be sent out after n seconds. (2) Solve the recurrence relation. (3) How many e-mails are sent at the end of 20 seconds

1.START T=0.....

1000 EMAILS SENT AND RECEIVED BY 1000 M/CS.

T=1.....

EACH OF THE M/C SENDS 10 NEW MAILS ....

.................................

LET M[N] BE THE NUMBER OF MAILS SENT OUT AFTER N SECONDS.

SO , EACH OF THESE M/CS WILL SEND 10 MAILS IN NEXT 1 SECOND.

HENCE NUMBER OF MAILS SENT IN N+1 SECONDS=M[N+1]=

M[N+1]=10*M[N].......................1

THIS IS THE RECURRENCE RELATION.....

2.SOLUTION .....

M[N+1]=10M[N]=10*10M[N-1]=10*10*10M[N-2]=........

M(N+1)=[10^1][M(N)]=[10^2][M(N-1)]=[10^3][M(N-2)]=..........=[10^N][M(1)]=[10^(N+1)][M(0)]

M[N+1]=[10^(N+1)][1000]=[10^(N+1)][10^3]=[10^(N+4)]......................................2

THIS IS THE NUMBER OF MAILS SENT AFTER N+1 SECONDS .....OR ....

M[N]=[10^(N+3)].............................................3

..................IS THE SOLUTION FOR NUMBER OF MAILS SENT AFTER N SECONDS.....

3.AFTER N=20 SECONDS , THE ANSWER IS ....

M[20]=10^(20+3)=10^23

8 0
3 years ago
Read 2 more answers
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