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Mnenie [13.5K]
3 years ago
15

Evaluating numeric expressions 16^5/4X16^1/4/(16^1/2)^2=?

Mathematics
2 answers:
Sever21 [200]3 years ago
7 0

Answer: 2^14

Step-by-step explanation:

miskamm [114]3 years ago
3 0

Answer:

its 4

Step-by-step explanation:

i just did the quiz

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Explain how to use mental math to express 12/25 as a percent
o-na [289]

Answer:

Step-by-step explanation:

12/25 =

12 ÷ 25 =

0.48 =

0.48 × 100/100 =

0.48 × 100% =

(0.48 × 100)% =

48%;

8 0
3 years ago
1. if x० and 70० makes an angle of complement, find the value of x.
stich3 [128]

Answer:

don't spend our time here

this is a spamm community

6 0
3 years ago
An office is divided into 12 cubicles. How many of the cubicles are carpeted if 1/3 of the cubicles are carpeted
11Alexandr11 [23.1K]
12÷3=4
cause 1/3 in this situation is 3 cause 12÷4 is 3
5 0
3 years ago
a 400 g of a liquid of density 1.6g/cm3 is made from mixing liquids of density 1.2g/cm3 and 1.8g/cm3 find the mass of the liquid
gogolik [260]

Answer:

133.33g

Step-by-step explanation:

Let the:

Mass of 1.2g/cm³ of liquid = x

Mass of 1.8g/cm³ of liquid = y

From our Question above, our system of equations is given as:

x + y = 400........ Equation 1

x = 400 - y

1.2 × x + 1.8 × y = 1.6 × 400

1.2x + 1.8y = 640..... Equation 2

We substitute, 400 - y for x in Equation 2

1.2(400 - y) + 1.8y = 640

480 - 1.2y + 1.8y = 640

- 1.2y + 1.8y = 640 - 480

0.6y = 160

y = 160/0.6y

y = 266.67 g

Solving for x

x = 400 - y

x = 400 - 266.67g

x = 133.33g

Therefore, the mass of the liquid of density 1.2g/cm³ is 133.33g

7 0
3 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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