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vladimir1956 [14]
3 years ago
7

there are 810 calories in 3 scoops of vanilla ice scream. How many calories are there in 7 scoops of ice screams

Mathematics
1 answer:
Eddi Din [679]3 years ago
7 0
810/3=270 You divide 810 by 3 to find the number of calories in 1 scoop
270*7=1890 You multiply 270*7 to find the total calories in 7 scoops
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What is the decimal equivalent of an 11% increase
valentinak56 [21]

Answer:

0.11

Step-by-step explanation:

You would convert the percentage to a fraction. Since the % symbol means 'out of 100', you would put 11 over 100: 11/100.

Then you would convert the fraction to a decimal: 0.11.

An easier way could also be to take the percentage and move it two decimal places to the right.

Ex: 43% moved two decimal places to the right would be 0.43.

I hope this helps!

3 0
3 years ago
Find M<1, M<2, M<3, M<4, M<5
Leokris [45]

Answer:

Step-by-step explanation:

m∠2 = 52° [Alternate interior angles]

m∠1 + m∠2 = 180° [Linear pair of angles are supplementary]

m∠1 = 180° - 52°

m∠1 = 128°

m∠3 = 47° [Alternate interior angles]

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m∠5 = 180° - 99°

        = 81°

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6 0
3 years ago
Which ratio expresses the scale used to create this drawing? The school has dimensions of 18 feet by 18 feet.
Papessa [141]
Answer A 1 square/2 feet=9 squares/18 feet
8 0
3 years ago
Read 2 more answers
A) Find a recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s.
Fed [463]

Answer:

A) a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

B) a_{0} = a_{1} = 0

C)   for n = 2

  a_{2} = 1

for n = 3

 a_{3} = 3

for n = 4

a_{4} = 8

for n = 5

a_{5} = 19

Step-by-step explanation:

A) A recurrence relation for the number of bit strings of length n that contain a  pair of consecutive Os can be represented below

if a string (n ) ends with 00 for n-2 positions there are a pair of  consecutive Os therefore there will be : 2^{n-2} strings

therefore for n ≥ 2

The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os

a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

b ) The initial conditions

The initial conditions are : a_{0} = a_{1} = 0

C) The number of bit strings of length seven containing two consecutive 0s

here we apply the re occurrence relation and the initial conditions

a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

for n = 2

  a_{2} = 1

for n = 3

 a_{3} = 3

for n = 4

a_{4} = 8

for n = 5

a_{5} = 19

7 0
3 years ago
Is -6.06 a natural number
Rus_ich [418]

Answer:

Natural numbers are all positive numbers like 1, 2, 3, 4, and so on. They are the numbers you usually count and they continue till infinity. Whereas, the whole numbers are all natural numbers including 0, for example, 0, 1, 2, 3, 4, and so on. Integers include all whole numbers and their negative counterpart.

Step-by-step explanation:

3 0
2 years ago
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