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Nana76 [90]
3 years ago
9

Given C is the midpoint of BD.

Mathematics
1 answer:
Nataly_w [17]3 years ago
4 0

Answer:definition of midpoint, angle BCA is congruent to angle DCA

You might be interested in
write a point-slope equation for the line that passes through the point (6,8) and is parallel to the line given by y=-5x+4
VashaNatasha [74]

Answer:

  y -8 = -5(x -6)

Step-by-step explanation:

The point-slope form of the equation for a line is generally written ...

  y -k = m(x -h)

for slope m and point (h, k).

The slope of your parallel line is the same as the slope of the reference line, -5. So your equation is ...

  y -8 = -5(x -6)

3 0
3 years ago
Find the nth taylor polynomial for the function, centered at c. f(x) = ln(x), n = 4, c = 5
masha68 [24]

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Given:

f(x) = ln(x)

n = 4

c = 3

nth Taylor polynomial for the function, centered at c

The Taylor series for f(x) = ln x centered at 5 is:

P_{n}(x)=f(c)+\frac{f^{'} (c)}{1!}(x-c)+  \frac{f^{''} (c)}{2!}(x-c)^{2} +\frac{f^{'''} (c)}{3!}(x-c)^{3}+.....+\frac{f^{n} (c)}{n!}(x-c)^{n}

Since, c = 5 so,

P_{4}(x)=f(5)+\frac{f^{'} (5)}{1!}(x-5)+  \frac{f^{''} (5)}{2!}(x-5)^{2} +\frac{f^{'''} (5)}{3!}(x-5)^{3}+.....+\frac{f^{n} (5)}{n!}(x-5)^{n}

Now

f(5) = ln 5

f'(x) = 1/x ⇒ f'(5) = 1/5

f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25

f'''(x) = 2/x³  ⇒ f'''(5) = 2/5³ = 2/125

f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625

So Taylor polynomial for n = 4 is:

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Hence,

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Find out more information about nth taylor polynomial here

brainly.com/question/28196765

#SPJ4

3 0
1 year ago
The pesticide diazinon is in common use to treat infestations of the German cockroach, Blattella germanica. A study investigated
cluponka [151]

Answer:

We conclude that there is no difference in the proportion of cockroaches that died on each surface.

Step-by-step explanation:

We are given that a study investigated the persistence of this pesticide on various types of surfaces.

After 14 days, they randomly assigned 72 cockroaches to two groups of 36, placed one group on each surface, and recorded the number that died within 48 hours. On the glass, 18 cockroaches died, while on plasterboard, 25 died.

<em>Let </em>p_1<em> = proportion of cockroaches that died on glass surface.</em>

<em />p_2<em> = proportion of cockroaches that died on plasterboard surface.</em>

So, Null Hypothesis, H_0 : p_1-p_2 = 0      {means that there is no difference in the proportion of cockroaches that died on each surface}

Alternate Hypothesis, H_A : p_1-p_2\neq 0      {means that there is a significant difference in the proportion of cockroaches that died on each surface}

The test statistics that would be used here <u>Two-sample z proportion</u> <u>statistics</u>;

                       T.S. =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}  } }  ~ N(0,1)

where, \hat p_1 = sample proportion of cockroaches that died on glass surface = \frac{18}{36} = 0.50

\hat p_2 = sample proportion of cockroaches that died on plasterboard surface = \frac{25}{36} = 0.694

n_1 = sample of cockroaches on glass surface = 36

n_2 = sample of cockroaches on plasterboard surface = 36

So, <u><em>test statistics</em></u>  =  \frac{(0.50-0.694)-(0)}{\sqrt{\frac{0.50(1-0.50)}{36}+\frac{0.694(1-0.694)}{36}  } }

                               =  -1.712

The value of z test statistics is -1.712.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that there is no difference in the proportion of cockroaches that died on each surface.

8 0
3 years ago
Is each line parallel, perpendicular, or neither parallel nor perpendicular to the line −3x+5y=−15? Drag and drop each choice in
nexus9112 [7]

The first line is perpendicular, the second line is not perpendicular and not parallel , the third line is parallel to the above line and the forth line not parallel or perpendicular to the line.

<h3>How can the slope of the line be calculate?</h3>

The slope of the line  is been given as −3x+5y=−15

Then y= \frac{3}{5x} -3, hence the slope is \frac{3}{5}

  • Slope of the first line 5x + 3y = 15 after calculation is

the slope is y= \frac{5}{-3x} +5\\\\\frac{5}{-3}

this means that the line is perpendicular to the given line

  • Slope of the second line  3x + 5y = 15 after calculation is

y= \frac{3}{-5x} +15\\\\\frac{3}{-5}

This means that the line is not perpendicular and not parallel

  • Slope of the Third line  -3x + 5y = 15 can be calculated as

y= \frac{3}{5x} +15\\\\\frac{3}{5}

This means that the line is parallel to the above line.

  • Slope of the first line  3x + 5y = 15 after calculation is

y= \frac{-3}{5x} +3\\\\\frac{-3}{5}

This means that the line is not parallel or perpendicular to the line

Learn more about slope of  line at:

brainly.com/question/3493733

#SPJ1

5 0
1 year ago
A certain scale has an uncertainty of 4 g and a bias of 2 g. A single measurement is made on this scale. What are the bias and u
Fed [463]

Answer: 0.8

Step-by-step explanation:

given data:

uncertainity = 4g

bias = 2g

solution:

σX =0.2 and σY = 0.4

σcX

= 3σX

= 4(0.2)

= 0.8

6 0
3 years ago
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