Which describes how triangle D was transformed to result in similar triangle D'?
2 answers:
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Answer:
x<-8 or x>1/5
Step-by-step explanation:
First we replace > symbol by = sign
To solve for x we multiply 5x-1 on both sides
x+8 =0
x=-8
5x-1=0 solve for x
x= 1/5
WE got two values x=-8 and x= 1/5
Now we make a number line and check with each interval
x<-8, -8<x<1/5, x>1/5
Pick x= -9 and check with the given inequality
1/46 >0 is true, so x<-8 satisfies our inequality
Pick x= 0 and check with the given inequality
-8 >0 is false , -8<x<1/5 does not satifies our inequality
Pick x= 2 and check with the given inequality
10/9>0 is true , x>1/5 satisfies our inequality
Answer:
a) For this case we define the random variable as X ="waiting time during peak hours" and we know that this distribution follows an uniform distribution:
Where a and b represent the limits of the distribution.
b)
And the height for this case would be 0.125
Step-by-step explanation:
Part a
For this case we define the random variable as X ="waiting time during peak hours" and we know that this distribution follows an uniform distribution:
Where a and b represent the limits of the distribution.
Part b
For this case the density function would be given by:
And the height for this case would be 0.125
And for other case.
The cumulative distribution function would be given by:
Answer:
The shadow is decreasing at the rate of 3.55 inch/min
Step-by-step explanation:
The height of the building = 60ft
The shadow of the building on the level ground is 25ft long
Ѳ is increasing at the rate of 0.24°/min
Using SOHCAHTOA,
Tan Ѳ = opposite/ adjacent
= height of the building / length of the shadow
Tan Ѳ = h/x
X= h/tan Ѳ
Recall that tan Ѳ = sin Ѳ/cos Ѳ
X= h/x (sin Ѳ/cos Ѳ)
Differentiate with respect to t
dx/dt = (-h/sin²Ѳ)dѲ/dt
When x= 25ft
tanѲ = h/x
= 60/25
Ѳ= tan^-1(60/25)
= 67.38°
dѲ/dt= 0.24°/min
Convert the height in ft to inches
1 ft = 12 inches
Therefore, 60ft = 60*12
= 720 inches
Convert degree/min to radian/min
1°= 0.0175radian
Therefore, 0.24° = 0.24 * 0.0175
= 0.0042 radian/min
Recall that
dx/dt = (-h/sin²Ѳ)dѲ/dt
= (-720/sin²(67.38))*0.0042
= (-720/0.8521)*0.0042
-3.55 inch/min
Therefore, the rate at which the length of the shadow of the building decreases is 3.55 inches/min
