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riadik2000 [5.3K]
3 years ago
8

A Web music store offers two versions of a popular song. The size of the standard version is 2.4 megabytes (MB). The size of the

high-quality version is 4.9 MB. Yesterday, there were 1020 downloads of the song, for a total download size of 3398 MB. How many downloads of the standard version were there?
Mathematics
1 answer:
ANEK [815]3 years ago
7 0

Answer:

50.mb

Step-by-step explanation:

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In ΔJKL, j = 74 cm, k = 14 cm and l=80 cm. Find the measure of ∠J to the nearest degree.
natita [175]

Answer:

  ∠J = 60°

Step-by-step explanation:

The Law of Cosines tells you ...

  j² = k² +l² -2kl·cos(J)

Solving for J gives ...

  J = arccos((k² +l² -j²)/(2kl))

  J = arccos((14² +80² -74²)/(2·14·80)) = arccos(1120/2240) = arccos(1/2)

  J = 60°

_____

<em>Additional comment</em>

It is pretty rare to find a set of integer side lengths that result in one of the angles of the triangle being a rational number of degrees.

6 0
2 years ago
Find the angles marked with letters. <br>Please help!! ​
Paladinen [302]
A=34°
B=76°
C and d = 70°
because a triangle is 180° and we know
a+b+c=180
76+34=110
180-110=70
3 0
2 years ago
Simply 3/4s - (6 - 3/4s)
Stolb23 [73]

Answer:  3/2 s-6

hope i helped

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8 0
3 years ago
Read 2 more answers
HELP<br> Write an equation of a line passing through (-8, 6) and parallel to the line y=-4x - 1
MArishka [77]

The line is parallel so it has the same slope as the other line, -4.

Substitute the x and y variables with the given points and find the y-intercept.

(-8, 6)

y = -4x + b

6 = -4(8) + b

6 = -32 + b

6 + 32 = b

38 = b

So the equation is:

y = -4 + 38

4 0
3 years ago
Jim wants to build a rectangular parking lot along a busy street but only has 2400 feet of fencing available. if no fencing is r
PSYCHO15rus [73]
<span>Length = 1200, width = 600
   First, let's create an equation for the area based upon the length. Since we have a total of 2400 feet of fence and only need to fence three sides of the region, we can define the width based upon the length as: W = (2400 - L)/2
   And area is: A = LW
    Substitute the equation for width, giving: A = LW A = L(2400 - L)/2
    And expand: A = (2400L - L^2)/2 A = 1200L - (1/2)L^2
    Now the easiest way of solving for the maximum area is to calculate the first derivative of the expression above, and solve for where it's value is 0. But since this is supposedly a high school problem, and the expression we have is a simple quadratic equation, we can solve it without using any calculus. Let's first use the quadratic formula with A=-1/2, B=1200, and C=0 and get the 2 roots which are 0 and 2400. Then we'll pick a point midway between those two which is (0 + 2400)/2 = 1200. And that should be your answer. But let's verify that by using the value (1200+e) and expand the equation to see what happens:
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 A = 1440000+1200e - 720000 - 1200e - (1/2)e^2
 A = 720000 - (1/2)e^2
   And notice that the only e terms is -(1/2)e^2. ANY non-zero value of e will cause this term to be non-zero and negative meaning that the total area will be reduced. Therefore the value of 1200 for the length is the best possible length that will get the maximum possible area.</span>
7 0
3 years ago
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