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olganol [36]
2 years ago
6

What is i hcf of 21 and 28

Mathematics
2 answers:
olchik [2.2K]2 years ago
7 0
21|3 \\ 7 \ |7 \\ 1 \\\\ 21=3*7 \\\\ 28|2 \\ 14|2 \\ 7 \ |7 \\ 1 \\\\ 28=2^2*7 \\\\ H.C.F(21;28)\to\boxed{7}
alexira [117]2 years ago
5 0
The factors of  21  are  1,  3,  7, and  21 .

The factors of  28  are  1,  2,  4,  7,  14, and  28 .

The factors common to both numbers are  1  and  7 .

The greatest one is <em> 7</em>.
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<img src="https://tex.z-dn.net/?f=y%3Dx%2B3y%3Dx-1" id="TexFormula1" title="y=x+3y=x-1" alt="y=x+3y=x-1" align="absmiddle" class
Dominik [7]

Answer:

y = x + 3y = x - 1

Step-by-step explanation:

Just graph them!

Hope I could help.

5 0
3 years ago
The odds of winning a rifle are 1:9 the probability of winning a carnival game is 0.15 does the raffle or the carnival game give
Lilit [14]

Answer:

Hence the carnival game gives you better chance of winning.

Step-by-step explanation:

Let the event of win be given by 1/10 in the game of rifle then the event of loose is given by 9/10

the

Odds in favor of a game are given by  = P(Event)/ 1- P(Event)

Odds in favor of winning a rifle are given by = 1/10/ 1- 1/10

                                                                         =1/10/9/10

                                                                          =1/9

                                                                             = 0.111

The probability of winning aa rifle game is 0.111

The probability of winning the carnival game is 0.15

Comparing the two probabilities   0.111:0.15

The probability of  winning carnival game is greater than winning a rifle game

0.15>0.11

Hence the carnival game gives you better chance of winning.

3 0
2 years ago
To determine the density of grains, a student uses a 50ml beaker graded by 5ml increments and a scale with 1g absolute uncertain
labwork [276]

Answer:

The mass of the grains = 120 ± 1 g

Step-by-step explanation:

we are given the following:

Total mass of container + grains = 185 grams

Mass of container = 65 grams

Therefore, mass of grains is calculated as follows:

Mass of grains = ( Mass of container + grains) - mass of container

= 185 - 65 = 120 grams.

since the scale has an absolute uncertainty of 1 g, the mass of the grains is written as 120 ± 1 g

8 0
3 years ago
Why is it advantageous to fill out the Budget and Cash Flow spreadsheet at the start of the simulation?
Tcecarenko [31]

Answer:

It is likewise significant on the grounds that it causes you decide if your business has enough cash to run or to grow it in future. Thus budget and cash flow spreadsheet is an absolute necessity in a simulation to grow.

Step-by-step explanation:

Cash flow spreadsheet alludes to the announcement of planned cash inflows and outflows. Budget cash flow spreadsheet is utilized to assess the momentary cash necessity and it can likewise be utilized to distinguish where the most extreme cash is going out and from where is the greatest inflow.

8 0
3 years ago
According to data from a medical​ association, the rate of change in the number of hospital outpatient​ visits, in​ millions, in
GaryK [48]

Answer:

a) f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient​ visits, in​ millions, is given by:

f'(t)=0.001155t(t-1980)^{0.5}

a) To find the function f(t) you integrate f(t):

\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt

To solve the integral you use:

\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}

Next, you replace in the integral:

\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C

Then, the function f(t) is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient​ visits.

Hence C' = 264,034,000

The function is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) For t = 2015 you have:

f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7

3 0
3 years ago
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