Given :
C, D, and E are col-linear, CE = 15.8 centimetres, and DE= 3.5 centimetres.
To Find :
Two possible lengths for CD.
Solution :
Their are two cases :
1)
When D is in between C and E .
. . .
C D E
Here, CD = CE - DE
CD = 15.8 - 3.5 cm
CD = 12.3 cm
2)
When E is in between D and C.
. . .
D E C
Here, CD = CE + DE
CD = 15.8 + 3.5 cm
CD = 19.3 cm
Hence, this is the required solution.
Answer:
My answer is 1170 but the way I figured out the problem was by listing numbers 1-30. the problem state that the 1st row had 10, 2nd row had 12, and 3rd row had 14 and so forth. So I basically did the same method till I got to 30 and then add up all the numbers which gave me the answer of 1170.
Step-by-step explanation:
For each, you'll use the slope formula
m = (y2-y1)/(x2-x1)
For function f, you'll use the two points (1,6) and (2,12) since x ranges from x = 1 to x = 2 for function f
The slope through these two points is
m = (y2-y1)/(x2-x1)
m = (12-6)/(2-1)
m = 6/1
m = 6
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For function g, you'll use (2,4) and (3,20)
The slope through these two points is
m = (y2-y1)/(x2-x1)
m = (20-4)/(3-2)
m = 16/1
m = 16
-------------------------------------------
For function h, you'll use (0,-6) and (2,-18). The y coordinates can be found by plugging in x = 0 and x = 2 respectively into h(x)
The slope through these two points is
m = (y2-y1)/(x2-x1)
m = (-18-(-6))/(2-0)
m = (-18+6)/(2-0)
m = (-12)/(2)
m = -6
-------------------------------------------
The order from left to right is: h, f, g
Answer:
See explanation
Step-by-step explanation:
Maximum weight = 20 kg
Bag 1 = 3.5kg
Bag 2 = 15 kg
Bag 3 = 2kg
Bag 4 = 1.5kg
Total weight of bags = 22 kg
Excess weight of his luggage = Total weight of bags - Maximum weight
= 22 kg - 20 kg
= 2 kg
Express the excess weight as a percentage of his maximum weight allowed = excess weight / maximum weight × 100
= 2/20 × 100
= 0.1 × 100
= 10%
Express the excess weight as a percentage of his maximum weight allowed = 10%
Answer:
144 ways
Step-by-step explanation:
Number of paintings = 7
Renaissance = 4
Baroque = 3
We are hanging from left to right and we will first hang Renaissance painting before baroque painting.
For Renaissance we have 4! Ways of doing so. 4 x3x2x1 = 24
For baroque we have 3! Ways of doing so. 3x2x1 = 6
We have 4!ways x 3!ways
= (4x3x2x1) * (3x2x1) ways
= 144 ways
Therefore we have 144 ways to hang the painting.
