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2 years ago

Lesson 22 exit ticket

Mathematics

1 answer:

natita [175]2 years ago

5 0

What about it. anything spesific?

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A baker sold 120 banana muffins and 60 blueberry muffins what was the ratio of banana muffins to blueberry muffin sold and

andrew-mc [135]

Answer:

120:60 or :=to

Step-by-step explanation:

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2 years ago

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Pls help me with this

jok3333 [9.3K]

Answer would be: B.) 3/8 cup

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3 years ago

Ten times as much as 3000

chubhunter [2.5K]

Multiply 3000 by 10

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3 years ago

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Today there are a total of six toll-free area codes: 800, 844, 855, 866, 877, and 888. Assume that all seven digits for the rest

Ludmilka [50]

Answer:

$6 \times 10^{7}$

Step-by-step explanation:

Total number of toll-free area codes = 6

A complete number will be of the form:

800-abc-defg

Where abcdefg can be any 7 numbers from 0 to 9. This holds true for all the 6 area codes.

Finding the possible toll free numbers for one area code and multiplying that by 6 will give use the total number of toll free numbers for all 6 area codes.

Considering: 800-abc-defg

The first number "a" can take any digit from 0 to 9. So there are 10 possibilities for this place. Similarly, the second number can take any digit from 0 to 9, so there are 10 possibilities for this place as well and same goes for all the 7 numbers.

Since, there are 10 possibilities for each of the 7 places, according to the fundamental principle of counting, the total possible toll free numbers for one area code would be:

Possible toll free numbers for 1 area code = 10 x 10 x 10 x 10 x 10 x 10 x 10 = $10^{7}$

Since, there are 6 toll-free are codes in total, the total number of toll-free numbers for all 6 area codes = $6 \times 10^{7}$

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3 years ago

Using a linear approximation, estimate f(2.1), given that f(2) = 5 and f'(x) = √3x-1.

algol [13]

Answer:

$f\left( {2.1} \right) \approx 5.22360$ .

Step-by-step explanation:

The linear approximation is given by the equation

${f\left( x \right) \approx L\left( x \right) }={ f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right).}$

Linear approximation is a good way to approximate values of $f(x)$ as long as you stay close to the point $x= a$ , but the farther you get from $x=a$ , the worse your approximation.