Answer:
EG = 19
Step-by-step explanation:
* Lets explain how to solve the problem
- If a line bisects another line that means the point of intersection
divides the second line into two equal parts
∵ EF bisects CD at G
∴ CG = GD
∵ CG = 5x - 1
∵ GD = 7x - 13
∴ 7x - 13 = 5x - 1
* Lets solve the equation
∵ 7x - 13 = 5x - 1
- Subtract 5x from both sides and add 13 to both sides
∴ 7x - 5x = 13 - 1
∴ 2x = 12
- Divide both sides by 2
∴ x = 6
- Point G divides EF into two parts EG and GF
∴ EF = EG + GF
∵ EF = 6x - 4
- Substitute the value of x to find EF
∵ x = 6
∴ EF = 6(6) - 4 = 36 - 4 = 32
∴ EF = 32
∵ GF = 13
- Substitute the values of EF and GF in the equation of EF
∴ 32 = EG + 13
- Subtract 13 from both sides
∴ 19 = EG
* EG = 19
Answer:
8 hot dogs
Step-by-step explanation:
Answer:
1:500000
Step-by-step explanation:
1 mm (map) equals 500 m (actual) .
Let's convert 500 m to mm.
1m = 1000mm
500m = 500000 mm
So 1mm to 500000mm on a scale is
1:500000
So it's all about converting the metre to million metre then doing the ratio.
In this case we are not to divide anything because it's already in 1.
So it's 1mm on paper then 500000mm on actual.
Thank you
Answer:
a) 0.2
b) 0.2
c) 0.5
Step-by-step explanation:
Let 
be the event "the car stops at the signal.
In the attached figure you can see a tree describing all possible scenarios.
For the first question the red path describes stopping at the first light but not stopping at the second. We can determine the probability of this path happening by multiplying the probabilities on the branches of the tree, thus
For the second one the blue path describes the situation
For the las situation the sum of the two green path will give us the answer
A divergent series is a type of series in which the ratio is too large and the summation is held from one to infinity. Hence, the sum could not be found out. A convergent series, on the other hand, has a definite answer because one could be the ratio is small and the lower and upper limits are defined not equal to infinity either positive or negative.
