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DENIUS [597]
3 years ago
9

Find the measure of angle JCA if J is the incenter of the triangle ABC.

Mathematics
1 answer:
kobusy [5.1K]3 years ago
7 0

let's recall that the incenter is where all angle bisectors for the triangle meet, and that an angle bisector cuts an angle into two equal halves.

Check the picture below.

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200/5 = 40...she should read at least 40 per day to reach her goal
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3 years ago
Can anyone help me please? I have no idea what to do! :'(
liberstina [14]

Answer:

7) 10^(3/2)

8) 2^(1/6)

9) 2^(5/4)

10) 5^(5/4)

Step-by-step explanation:

7)  (√10)^3 = 10^(3/2)

8)  6 root 2 = 2^(1/6)

9)  (4 root 2)^5 = 2^(5/4)

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5 0
3 years ago
1/2x +1/4y =32 , then find 2x+y
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Answer:

128

Step-by-step explanation:

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3 years ago
A study of the relationship between age and various visual functions (such as acuity and depth perception) reported the followin
maks197457 [2]

Answer:

(a) \sum x_i=56.68 and \sum x_i^2=197.46

(b) The sample variance is s^2=0.530 and the sample standard deviation is s=0.728

Step-by-step explanation:

(a)

The sum of these 17 sample observations is

\sum x_i=2.79\:+2.57+\:2.73\:+3.75\:+2.27+\:2.75+\:4.00+\:4.22+\:3.88+\:4.35+\:3.41+\:4.57+\:2.38+\:3.73\:+2.75+\:3.47+\:3.06\\\\\sum x_i=56.68

and the sum of their squares is

\sum x_i^2=2.79^2\:+2.57^2+\:2.73^2\:+3.75^2\:+2.27^2+\:2.75^2+\:4.00^2+\:4.22^2+\:3.88^2+\:4.35^2+\:3.41^2+\:4.57^2+\:2.38^2+\:3.73^2\:+2.75^2+\:3.47^2+\:3.06^2\\\\\sum x_i^2=197.46

(b)

The sample variance, denoted by s^2, is given by

s^2=\frac{S_{xx}}{n-1}

where S_{xx} =\sum x_i^2-\frac{(\sum x_i)^2}{n}

Applying the above formula we get that

S_{xx}=197.46-\frac{(56.68)^2}{17}\\\\S_{xx} =8.482

s^2=\frac{8.482}{17-1}=0.530

The sample standard deviation, denoted by <em>s</em>, is the (positive) square root of the variance:

s=\sqrt{s^2}

Applying the above formula we get that

s=\sqrt{0.530}=0.728

3 0
3 years ago
1. algebraic expression
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hope this helps
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