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Alexus [3.1K]
3 years ago
14

If H is the circumcenter of triangle BCD, find each measure. CD, CE, HD, GD, HG, HF

Mathematics
1 answer:
Levart [38]3 years ago
7 0

Answer:

H is the circumcenter of triangle BCD

  • CD = 2FD = 2*32 = 64
  • CE = BE = 26
  • HD = HC = 33
  • GD = 1/2BD = 1/2(58) = 29
  • HG = √HD ²-GD² = √33²-29² = 2√62 ≈ 15.75
  • HF = √HD ²-FD² = √33²-32² =√65 ≈ 8.06
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Evaluate the expression if y=4.4 and z= 12<br> z2+ 8y
Evgen [1.6K]

Answer: <em>59.2</em>

Step-by-step explanation:

<em>Take your equation</em>

<em>z2+8y</em>

<em>Now plug in the values given</em>

<em>(12)2+8(4.4)</em>

<em>24+35.2</em>

<em>59.2</em>

5 0
3 years ago
round 205,721 to the nearest hundred thousands , to the nearest ten thousands and to the nearest thousands
Dovator [93]
To the nearest hundred thousandths: 200,000
To the nearest ten-thousandths: 210,000
To the nearest thousands: 206,000
7 0
3 years ago
Find C to form completing Square<br> <img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B5x%20%2B%20C" id="TexFormula1" title="x^
postnew [5]

Answer:

\frac{25}{4}  (or 6.25)

Step-by-step explanation:

Let's do a simple square problem to see the relationship between the elements of the resulting equation...

(x+3)(x+3) = x(x+3) + 3(x+3) = x^{2} +3x + 3x + 9 = x^{2} + 6x + 9

As you see, the number of 'x' (6) is the double of the numeric part of the binomial expression (3), while the numeric-only part (9) is the square of the numeric part of the binomial expression (3).

So, if we look at the formula from the problem, we see the the number of 'x' is 5. To obtain the C value, we need to divide the 5 by 2... then get the square value of the result.

\frac{5}{2} * \frac{5}{2} = \frac{25}{4}  (or 6.25)

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3 years ago
In a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Interne
Dvinal [7]

Answer:

b. Do not reject H0. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.7 hours.

Step-by-step explanation:

Given that in a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week. The mean of the sample observations was 12.9 hours.

H_0: \bar x =12.7\\H_a: \bar x >12,7

(Right tailed test at 5% level)

Mean difference = 0.2

Std error = \frac{6}{\sqrt{1000} } \\=0.1897

Z statistic = 1.0540

p value = 0.145941

since p >alpha we do not reject H0.

b. Do not reject H0. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.7 hours.

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3 years ago
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3 years ago
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