Is there a formula for problems like these? How do I even begin to solve it?

Mathematics

1 answer:

Pachacha [2.7K]3 years ago

5 0

Well, they're saying 9 with an exponet of 2 multiplied by something equals 27..... Hmmm.....

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X^2+y^2+14x+10y=7 <br> center and radius

Sholpan [36]

$(x-a)^2+(y-b)^2=r^2\\\\(a;\ b)-the\ coordinates\ of\ the\ center\\r-the\ radius\\\\==================================\\Use:(k+l)^2=k^2+2kl+l^2\ (*)\\==================================$

$x^2+y^2+14x+10y=7\\\\x^2+14x+y^2+10y=7\\\\x^2+2x\cdot7+y^2+2y\cdot5=7\\\\\underbrace{x^2+2x\cdot7+7^2}_{Use\ (*)}-7^2+\underbrace{y^2+2y\cdot5+5^2}_{Use\ (*)}-5^2=7\\\\(x+7)^2-49+(y+5)^2-25=7\\\\(x+7)^2+(y+5)^2-74=7\ \ \ \ |add\ 74\ to\both\ sides\\\\(x+7)^2+(y+5)^2=81\\\\(x+7)^2+(y+5)^2=9^2$

$Answer:\\\boxed{(-7;-5)-center;\ 9-radius}$

8 0

3 years ago

Read 2 more answers

a major metroplitan newspaper selected a simple random sample of 1,600 readers from their list of 100,000 subscribers. they aske

Arisa [49]

Answer:

The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.