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Jobisdone [24]
2 years ago
10

Help with evaluate and graph PLEASE HURRY

Mathematics
1 answer:
elixir [45]2 years ago
3 0

Answer:

Box 1 = -1

Box 2 = 8

Box 3 = 14

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Is this true in maths if a ate many sugar will a get diabites ? remider math question
lana66690 [7]
In math (or English, for that matter), a question is never true or false. Only a statement can have such attributes.

If you make the statement "if A ate many sugar, A will get diabetes," in math it cannot be decided wheter it is true or false without additional information about the truth values of the statements "A ate many sugar" and "A will get diabetes".
7 0
3 years ago
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=
ivann1987 [24]

9514 1404 393

Answer:

  • large: 55 lb
  • small: 30 lb

Step-by-step explanation:

Let x and y represent the weights of the large and small boxes, respectively. The problem statement gives rise to the system of equations ...

  x + y = 85 . . . . . combined weight of a large and small box

  70x +50y = 5350 . . . . combined weight of 70 large and 50 small boxes

We can subtract 50 times the first equation from the second to find the weight of a large box.

  (70x +50y) -50(x +y) = (5350) -50(85)

  20x = 1100 . . . . simplify

  x = 55 . . . . . . . divide by 20

Using this in the first equation, we can find the weight of a small box.

  55 +y = 85

  y = 30 . . . . . . . subtract 55

A large box weighs 55 pounds; a small box weighs 30 pounds.

4 0
2 years ago
2x - 5y = - 15
UkoKoshka [18]

Answer:

2x+15=5y

y=(2x+15)/5

hope this helps

7 0
2 years ago
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Monica measures 91 milliliters of water into 9 tiny beakers. she measures an equal amount of water into the first 8 beakers. she
UkoKoshka [18]
To determine the amount of water in the first 8 beakers, you will subtract the 19 ml that are in the 9th beaker from the total to see how much water was actually put into the first 8 beakers.

91 ml - 19 ml = 72 ml.

The 72 ml are divided evenly between 8 beakers.

72/8 = 9 ml

There would be 9 ml of water in each of the first 8 beakers.
8 0
3 years ago
5. I, II, III, IV, V,<br> Conjecture:
NemiM [27]
The conjecture is each term is the corresponding Roman numeral of the precious term
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3 years ago
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