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Ipatiy [6.2K]
4 years ago
15

A.51.25 B.51.15 C.50.85 D.50.45

Mathematics
1 answer:
irakobra [83]4 years ago
3 0

Answer:

maybe the answer is 51.25

Step-by-step explanation:

if you're not sure you have to figure out the mean

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Solve for x: 2x + 13 = 8x - 41
kherson [118]

Answer:

9 is your answer for x.

Step-by-step explanation:

Note the equal sign, what you do to one side, you do to the other. Isolate the variable, x. Do the opposite of PEMDAS.

PEMDAS is the order of operations, & =

Parenthesis

Exponent (& Roots)

Multiplication

Division

Addition

Subtraction

First, subtract 13 & 8x from both sides:

2x (-8x) + 13 (-13) = 8x (-8x) - 41 (-13)

2x - 8x = -41 - 13

Simplify. Combine like terms:

-6x = -54

Isolate the variable, x. Divide -6 from both sides:

(-6x)/-6 = (-54)/-6

x = (-54)/(-6)

x = 9

x = 9 is your answer.

~

8 0
3 years ago
Read 2 more answers
Use the distributive property to write an equivalent expression 4(5+19)=
Effectus [21]

Answer:

96

Step-by-step explanation:

4 ( 5 + 19 )

= 4 * 5 + 4 * 19

= 20 + 76

= 96

7 0
3 years ago
Read 2 more answers
We put 200 balls into 100 boxes such that every box got at least 1 ball and at most 100 balls. Prove that there are some boxes t
Vinvika [58]

Explanation:

Lets first show that al least 50 of the boxes contail at most 2 balls.

If there were 50 + k boxes with 3 balls or more, then we should have 100 - (50 + k) = 50 - k balls with 1 ball or 2. However in those 50 + k boxes with 3 or more balls we have alredy at least 3*(50+k) = 150 + 3k balls in them, and the amount of balls remaining is, as a result, at most 200 - (150 + 3k) = 50 - 3k, which cant be fit in 50 - k balls if we put at least 1 on each.

Therefore, there are at least 50 boxes with 1 or 2 balls. Whithin those boxes, we can obtain any number of balls selecting the appropiate boxes. Lets assume that we want M balls, and we have A boxes with 2 balls and B boxes with 1 ball, we have this possibilities (M equal or less than 2A + B, the total number of balls):

  • If M > 2A, then we pick all boxes with 2 balls (A in total) and M - 2A boxes with 1 ball. We have 2*A + (M-2A) = M. We are able to pick M - 2A boxes because B ≥ M - 2A.
  • If M ≤ 2A, and it is even, then we pick M/2 boxes with 2 balls.
  • If M ≤ 2A and it is odd, then we pick (M-1)/2 boxes with 2 balls and 1 box with 1 ball (if all boxes contain 2 balls or more, then we could pick 50 boxes with 2 balls because at least 50 boxes contain 1 or 2 balls; so we can assume that at least 1 box contain one single ball).

Lets call C the sum of the balls in the boxes with 1 or 2 balls. C should be at least 50. The argument made previously shows that we can pick boxes of 1 or 2 balls that cover any number of balls below to C. This means that we can obtain any number below 50; furthermore, if C is equal or greater than 100, then the problem is alredy solved. Lets suppose that C is lower than 100.  This means that the other boxes contain more than 100 balls in total.

Since we cant put more than 100 balls in one single box, then there should be a combination of boxes with 3 or more balls that contain between 50 and 100 balls. If that is not the case, then lets call L the biggest number of balls below 50 that we can obtain with boxes with 3 or more balls. Since the sum of all balls is bigger than 100, then there should be a box outside those we use to obtain L with 3 or more balls. Since L was the biggest number we could obtain below 50, and we are supposing that we cant obtain any number between 50 and 100, then that box should have more than 50 balls. Which means that that box alone could be used to obtain a number between 50 and 100. This is a contradiction.

The paragraph above shows that we can make a combination of boxes with 3 or more balls which combined number of balls is a number N between 50 and 100. Since we can make any number between 0 and 50 with boxes, for example 100 - N, with boxes of 1 or 2 balls, then we should be able to make exactly 100 balls using the boxes we have available.

I hope that works for you!

4 0
3 years ago
Complete the proof by providing the missing statement and reasons
Lisa [10]

Answer:

In triangle SHD and triangle STD.

\overline{SD} \perp \overline{HT}          [Side]

Since, a line is said to be perpendicular to another line if the two lines intersect at a right angle.

⇒ \angle SDH = \angle SDT = 90^{\circ}

\overline{SH} \cong \overline{ST}       [leg]               [Given]

Reflexive property states that the value is equal to itself.

\overline{SD} \cong \overline{SD}       [Leg]       [Reflexive property]

HL(Hypotenuse-leg) theorem states that any two right triangles that have a congruent hypotenuse and a corresponding congruent leg are the congruent triangles.


then, by HL theorem;

\triangle SHD \cong \triangle STD                       Proved!



6 0
4 years ago
The music center had 200 instruments. 70% of them were violins and the remaining are guitars. After a large sale of violins, the
belka [17]

Answer:

50 violins were  sold

Step-by-step explanation:

8 0
3 years ago
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