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svet-max [94.6K]
3 years ago
9

Select the three situations that model a sum of zero.

Mathematics
1 answer:
kaheart [24]3 years ago
4 0

Answer:

use the zero as an common multipal

Step-by-step explanation:

You might be interested in
PLEASE HELP
Pavel [41]

Answer:

15.06

Step-by-step explanation:

I don't have a step-by-step explanation, but i hope this helps!!!

7 0
3 years ago
Two sisters decide to take a series of acting lessons. One decides to pay a flat rate of $75, shown as f(x). The other just want
Mandarinka [93]
The answer is f(x) = 75, g(x) = 5x, h(x) = 5x + 75

Let x be the number of lessons.

The <u>flat rate of $75</u> as function f(x) can be expressed as: f(x) = <u>75</u>
<u>$5.00 per lesson x</u> as function g(x) can be expressed as: g(x) = <u>5x
</u>
The combination of two functions (f(x) and g(x)) is the amount the parent should pay:
h(x) = g(x) + f(x)
h(x) = 5x + 75
3 0
3 years ago
Read 2 more answers
53 gal 1 qt + 12 gal 3 qt + 82 gal 1 qt = ?​
mario62 [17]

Given:

53 gal 1 qt + 12 gal 3 qt + 82 gal 1 qt

To find:

The sum of the given expression.

Solution:

Let us first add gallons:

53 gal + 12 gal + 82 gal = 147 gal

Now, add quarts.

1 qt + 3 qt + 1 qt = 5 qt

We know that,

4 quarts = 1 gallon

So add 1 gallon to 147 gal.

147 gal + 5 qt = 147 gal + 4 qt + 1 qt

                      = 147 gal + 1 gal + 1 qt

                      = 148 gal + 1 qt

                      = 148 gal 1 qt

Therefore, 53 gal 1 qt + 12 gal 3 qt + 82 gal 1 qt = 148 gal 1 qt

4 0
3 years ago
Read 2 more answers
Please help with my geometry
erma4kov [3.2K]
25 = hypothenuse
14 = opposite
Use SOHCAHTOA
Give O and H, you can use sin
Sin^-1 = 14/25 = 34.0557
Round to hundreds
Solution: 34.06
8 0
3 years ago
How many different 7-place license plates are possible when 3 of the entries are letters and 4 are digits? Assume that repetitio
timofeeve [1]

Answer:

There are 6,151,600,000 different 7-place license plates are possible when 3 of the entries are letters and 4 are digits,

Step-by-step explanation:

For each of the entries which are letters, there are 26 possible outcomes.

For each of the entries which are digits, there are 10 possible outcomes.

These outcomes can be permutated.

For example, ABC1234 is a different outcome than A1B2C34. This means that we need to use the permutations formula.

The number of permutations of n, divided into two groups of size a and b, is:

P_{a,b}^{n} = \frac{n!}{a!b!}.

In this problem, we have a permutation of 7, divided into a group of 4(digits) and 3(letters).

How many different 7-place license plates are possible when 3 of the entries are letters and 4 are digits?

This is P_{3,4}^{7}*(26)^{3}*10^{4} = 35*(26)^{3}*10^{4} = 6,151,600,000

There are 6,151,600,000 different 7-place license plates are possible when 3 of the entries are letters and 4 are digits,

7 0
3 years ago
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