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Anna71 [15]
3 years ago
9

8. Define transversal lines.

Mathematics
1 answer:
balu736 [363]3 years ago
3 0

Step-by-step explanation:

In geometry, a transversal is a line that intersects two or more other (often parallel ) lines. In the figure below, line n is a transversal cutting lines l and m . When two or more lines are cut by a transversal, the angles which occupy the same relative position are called corresponding angles

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Find the surface area of a cone with radius 14.2 inches and slant height 24.1 inches.
ololo11 [35]

Answer:

1708.58658058

Step-by-step explanation:

Total surface area formula = (π×r²)+(π×r×s)

r = radius

s = slant height

So we substitute the values into the formula (π×r²)+(π×r×s)

(π×14.2²)+(π×14.2×24.1) = 1708.58658058

8 0
3 years ago
Can someone help me
algol13

Answer:

Step-by-step explanation:

A would be your answer.

3 0
3 years ago
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What is the equation of the line that passes through the point (1, 3) and is perpendicular to the line 2x+3y=15?
cricket20 [7]

Answer:

y = 1.5x + 1.5

Step-by-step explanation:

First, plot 2x + 3y = 15 on a graph. Then plot (1,3) on a graph. After that, draw a line that passes through (1,3) and forms 90 degrees angles with 2x + 3y = 15. Then find the equation of that line by finding the y-intercept (0, 1.5). Then head up to the intersection point (1.6153846153846, 3.9230769230769). Subtract 1.5 from y. (2.4230769230769) Then divide by 1.5 (1.6153846153846). Reverse the operations you just did (in reverse order) to get the equation y = 1.5x + 1.5.

(I don't know how to explain this too well. Here's a graph to make more sense of it.)

5 0
3 years ago
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Write 125.06 million 4 different ways.
madreJ [45]
125,060,000
125 and 06/100 million
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125 million, sixty thousand.
6 0
3 years ago
Read 2 more answers
We select n + 1 different integers from the set { 1 , 2 , ··· , 2 n } . Provethat there will alwaysbe two among the selected inte
just olya [345]

Answer:

See answer below

Step-by-step explanation:

From the set

{1,2,3,4...2n} we have 2n numbers in total , n are odd and n are even , therefore for a sample of n+1 numbers , we have at least 1 even number and 1 odd number.

Then

it the set includes 1 , the largest common divisor is 1 for 1 and the other numbers

if the set includes 3, there will be always a number that is not divisible by 3. Even we construct a set of n+1 numbers that are multiple of 3 , the largest number would be 3*(n+1)= 3*n+3 > 2*n (out of bounds) , therefore we are forced to take other number that is not divisible by 3  → the largest common divisor of that number with 3 is 1

If the set includes any other prime number → the largest common divisor of that with any other is 1

For the remaining odd numbers N, they can be factorised into other 2 odd common divisors N₂ and n₂ :

N = N₂*n₂ , since n₂ ≥ 2 →  N₂ < N

then the even N₂ also should be contained in the set

therefore also for N₂

N = N₃*n₃ →  N₃ < N₂

therefore if we continue , we would obtain a number  even Nn that has no smaller common divisors → since we cannot take all the multiples of N min ( because Nmin*(n+1)= Nmin*n+Nmin > 2*n for Nmin≥2) → there is at least a number in the sample of n+1 integers whose largest common divisor is 1

6 0
3 years ago
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