the equation in the slope-intercept of the side of triangle ABC that is perpendicular to segment EF is y = x + 1
<h3>How to determine the equation</h3>
From the figure given, we can deduce the coordinates of the sides
For A
A ( 4,2)
For B
B ( 4, 5)
C ( 1, 2)
D ( 2, -4 )
E ( 5, -4)
F ( 2, -1)
The slope for BC
Slope = 
Substitute the values for both B and C coordinates, we have
Slope = 
Find the difference for both the numerator and denominator
Slope = 
Slope = 1
We have the rotation for both point ( 0, 1)
y - y1 = m ( x - x1)
The values for y1 and x1 are 1 and 0 respectively and the slope m is 1
Substitute the values
y - 1 = 1 ( x - 0)
y - 1 = x
Make 'y' the subject of formula
y = x + 1
Thus, the equation in the slope-intercept of the side of triangle ABC that is perpendicular to segment EF is y = x + 1
Learn more about linear graphs here:
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W + 2w = 60
this equation is wrong
Answer:
Kindly check explanation
Step-by-step explanation:
Given the following :
Group 1:
μ1 = 59.7
s1 = 2.8
n1 = sample size = 12
Group 2:
μ2 = 64.7
s2 = 8.3
n2 = sample size = 15
α = 0.1
Assume normal distribution and equ sample variance
A.)
Null and alternative hypothesis
Null : μ1 = μ2
Alternative : μ1 < μ2
B.)
USing the t test
Test statistic :
t = (m1 - m2) / S(√1/n1 + 1/n2)
S = √(((n1 - 1)s²1 + (n2 - 1)s²2) / (n1 + n2 - 2))
S = √(((12 - 1)2.8^2 + (15 - 1)8.3^2) / (12 + 15 - 2))
S = 6.4829005
t = (59.7 - 64.7) / 6.4829005(√1/12 + 1/15)
t = - 5 / 2.5108165
tstat = −1.991384
Decision rule :
If tstat < - tα, (n1+n2-2) ; reject the Null
tstat < t0.1,25
From t table :
-t0.1, 25 = - 1.3163
tstat = - 1.9913
-1.9913 < - 1.3163 ; Hence reject the Null
22 and 27 is the answer. what else do you need??
I'm assuming that the 1/3 is an exponent.
If so, then
![2^{1/3} = \sqrt[3]{2}](https://tex.z-dn.net/?f=2%5E%7B1%2F3%7D%20%3D%20%5Csqrt%5B3%5D%7B2%7D)
Which is the cube root of 2. Raising any value to the 1/3 power is the same as taking the cube root.
