Using the given information, the midpoint of CD is (2.5, 4.5)
<h3>Midpoint of a line </h3>
From the question, we are to determine the midpoint of line CD
The midpoint of a line is given by the formula,
[(x₁ + x₂)/2 , (y₁ + y₂)/2]
From the given information, we are to find the midpoint of line CD with coordinates C(4,5) and D(1,4)
∴ x₁ = 4
x₂ = 1
y₁ = 5
y₂ = 4
Putting the parameters into the formula for midpoint, we get
[(x₁ + x₂)/2 , (y₁ + y₂)/2]
[(4 + 1)/2 , (5 + 4)/2]
(5/2 , 9/2)
(2.5, 4.5)
Hence, the midpoint of CD is (2.5, 4.5)
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Answer:
Step-by-step explanation:
The given ratio is 
9 is the common factor of 18 and 27
divide numerator and denominator by 9
hence is proportion with
The answer can be readily calculated using a single variable, x:
Let x = the amount being invested at an annual rate of 10%
Let (8000 - x) = the amount being invested at an annual rate of 12%
The problem is then stated as:
(x * 0.10) + ((8000 - x) * 0.12) = 900
0.10(x) + ((8000 * 0.12) - 0.12(x)) = 900
0.10(x) + 960 - 0.12(x) = 900
0.10(x) - 0.12(x) = 900 - 960
-0.02(x) = -60
-0.02(x) * -100/2 = -60 * -100/2
x = 6000 / 2
x = 3000
Thus, $3,000 is invested at 10% = $300 annually; and $8,000 - $3,000 = $5,000 invested at 12% = $600 annually, which sum to $900 annual investment.
Answer:
9/5 (K-273.15) + 32=F
Step-by-step explanation:
K=5 /9(F−32)+273.15
Subtract 273.15 from each side
K-273.15=5/9(F−32)+273.15-273.15
K-273.15=5/9(F−32)
Multiply by 9/5 on each side
9/5 (K-273.15)= 9/5 *5/9(F−32)
9/5 (K-273.15)=(F−32)
Add 32 to each side
9/5 (K-273.15) + 32=F−32 +32
9/5 (K-273.15) + 32=F
9514 1404 393
Answer:
- late only: 15
- extra-late only: 24
- one type: 43
- total trucks: 105
Step-by-step explanation:
It works well when making a Venn diagram to start in the middle (6 carried all three), then work out.
For example, if 10 carried early and extra-late, then only 10-6 = 4 of those trucks carried just early and extra-late.
Similarly, if 30 carried early and late, and 4 more carried only early and extra-late, then 38-30-4 = 4 carried only early. In the attached, the "only" numbers for a single type are circled, to differentiate them from the "total" numbers for that type.
__
a) 15 trucks carried only late
b) 24 trucks carried only extra late
c) 4+15+24 = 43 trucks carried only one type
d) 38+67+56 -30-28-10 +6 +6 = 105 trucks in all went out
