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TEA [102]
3 years ago
6

Ping's Ice Cream Palace offers a special sundae that contains over 2 kilograms of ice cream plus any number of toppings. Write a

n inequality that describes S, the weight of the special sundae in kilograms.
Mathematics
2 answers:
hjlf3 years ago
7 0

Let "x" represent the weight of the toppings. We know that we can have any number of toppings. This means that one may ask for no toppings at all too.

Now, we have been told that "S" is the weight of the special sundae in kilograms. This definitely included the "mandatory" 2 kilograms of ice cream. Therefore, S will be at-least equal to 2.

Thus, the inequality that describes S, the weight of the special sundae in kilograms at Ping's Ice Cream Palace is given as:

S\geq 2+x kilograms.

It can be seen that as x increases, S increases too and if an order does not want any toppings in it then the weight of the special sundae will be a minimum of 2 kg which is the weight of the ice cream.

julsineya [31]3 years ago
5 0

Answer:

The answer is S > 2

Step-by-step explanation:

I did the quiz.

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Answer:

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Which property is demonstrated below? b.1=b A. associative property B. distributive property C. commutative property D. identity
nydimaria [60]

Answer:

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Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Angelina wins money in a competition. She is given the choice as to how she is paid.
NISA [10]

Answer:

  • Choice 1
  • 20 days

Step-by-step explanation:

You are being asked to compare the value of a growing infinite geometric series to a fixed constant. Such a series will always eventually have a sum that exceeds any given fixed constant.

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<h3>a)</h3>

Angelina will get more money from the Choice 1 method of payment. The sequence of payments is a (growing) geometric sequence, so the payments and their sum will eventually exceed the alternative.

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<h3>c)</h3>

For a first term of 1 and a common ratio of 2, the sum of n terms of the geometric series is given by ...

  Sn = a1×(r^n -1)/(r -1) . . . . . . . . . . series with first term a1, common ratio r

We want to find n such that ...

  Sn ≥ 1,000,000

  1 × (2^n -1)/(2 -1) ≥ 1,000,000

  2^n ≥ 1,000,001 . . . . add 1

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8 0
3 years ago
Please answer this I need help asap
Yanka [14]

Answer:

(12, 48 )

Step-by-step explanation:

Given the 2 equations

y = 5x - 12 → (1)

y = 3x + 12 → (2)

Substitute y = 5x - 12 into (2)

5x - 12 = 3x + 12 ( subtract 3x from both sides )

2x - 12 = 12 ( add 12 to both sides )

2x = 24 ( divide both sides by 2 )

x = 12

Substitute x = 12 into either of the 2 equations and evaluate for y

Substituting into (2)

y = 3(12) + 12 = 36 + 12 = 48

solution is (12, 48 )

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bezimeni [28]

Answer:

F=ma

F=22.4*1.40=31.36

7 0
3 years ago
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