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AveGali [126]
3 years ago
10

The equation of s circle is x2+y2+6x+4y+10=1. What is this equation written in standard form?

Mathematics
1 answer:
Andreas93 [3]3 years ago
4 0

Answer:

i think it would be x two+ y two + six x + four y+ ten= one is that one the answers if so i hope this helps if not just give me the answers and i will try to help you as much as i can thanks

Step-by-step explanation:


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where does the graph of the function y=tan(x) have asymptotes? at the values of x where cos(x)=0 at the values of x where sin(x)
Gnom [1K]
Answer: at the values where cos(x) = 0

Justification:

1) tan(x) = sin(x) / cos(x).

2) functions have vertical asymptotes at x = a if Limit of the function x approaches a is + or - infinity.

3) the limit of tan(x) approaches +/- infinity where cos(x) approaches 0.

Therefore, the grpah of y = tan(x) has asymptotes where cos(x) = 0.

You can see the asympotes at x = +/- π/2 on the attached graph. Remember that cos(x) approaches 0 when x approaches +/- (n+1) π/2, for any n ∈ N, so there are infinite asymptotes.



7 0
3 years ago
Read 2 more answers
PLEAAASE HELP! i will manifest good things for you. i will mark u brainliest LOL
melamori03 [73]

Answer:

(c) is m=1 i am working on the other ones

5 0
3 years ago
Find the unpaid balance on the debt after 5 years of monthly payments on $190,000 at 3% for 25 years
NNADVOKAT [17]

Answer:

$114,000

Step-by-step explanation:

Use formula

I=P\cdot r\cdot t,

where

I = interst

P = principal

r = rate

t = time

First, find the interst for 25 years:

I = unknown

P = $190,000

r = 0.03 (3% as decimal)

t = 25

I_{25}=190,000\cdot 0.03\cdot 25=142,500

Now find the interest for 5 years:

I = unknown

P = $190,000

r = 0.03 (3% as decimal)

t = 5

I_5=190,000\cdot 0.03\cdot 5=28,500

The unpaid balance is

I=I_{25}-I_5=142,500-28,500=114,000

6 0
3 years ago
Factor the expression completely by first taking out a common factor<br> 3x^2 + 27x +24
Mrac [35]
The answer to this is
3(x+1)(x+8)

4 0
3 years ago
Suppose that administrators at a large urban high school want to gain a better understanding of the prevalence of bullying withi
boyakko [2]

Answer:

lower limit =0.261

upper limit =0.369

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Description in words of the parameter p

p represent the real population proportion of people that have experienced bullying

X= 63 people in the random sample that have experienced bullying

n=200 is the sample size required  

\hat p=\frac{63}{200}=0.315 represent the estimated proportion of people that have experienced bullying

z_{\alpha/2} represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Numerical estimate for p

In order to estimate a proportion we use this formula:

\hat p =\frac{X}{n} where X represent the number of people with a characteristic and n the total sample size selected.

\hat p=\frac{63}{200}=0.315 represent the estimated proportion of people that they were planning to pursue a graduate degree

Confidence interval

The confidence interval for a proportion is given by this formula  

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

For the 90% confidence interval the value of \alpha=1-0.90=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.64  

And replacing into the confidence interval formula we got:  

0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.261  

0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.369  

And the 90% confidence interval would be given (0.261;0.369).  

We are confident at 90% that the true proportion of people that they were planning to pursue a graduate degree is between (0.261;0.369).  

lower limit =0.261

upper limit =0.369

5 0
3 years ago
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