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jeyben [28]
3 years ago
8

Which of these phrases contain a variable? Check all that apply. A three cups of salsa B the weight of the puppy C the cost the

of dinner D twenty feet of wire
Mathematics
1 answer:
Lyrx [107]3 years ago
4 0

Answer:

phrases B and C are variables.

Step-by-step explanation:

Any thing whose value can be changed is called variable.

Whereas, if the value is fixed then it is called a constant.

We can use these definitions to find the phrases that contains a variable.

First option is: A three cups of salsa - It is constant because, the number of cups of salsa is fixed which is 3.

Second option is: the weight of the puppy - It is variable because, the weight of the puppy can be changed and not always fixed.

Third option is : the cost the of dinner- It is variable because, the cost of the dinner can be changed and depends on the quantity of the food ordered.

Last option: twenty feet of wire: It is a constant because the length of the wire is fixed which is 20 feet.

Therefore, the phrases B and C are variables.

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!!!!!!!!PLEASE HELP NOW !!!!!!!!!!!!!!!!!!
vampirchik [111]

Answer:

7

Step-by-step explanation:

You can convert the fourth square roots to 7^{\frac{1}{4}}} * 7^{\frac{1}{4}}} * 7^{\frac{1}{4}}} * 7^{\frac{1}{4}}}. Using the product of powers rule, we can add the four terms' exponents, resulting in 7^1, which is 7.

8 0
3 years ago
Please help me with this question!
frez [133]
\text {Laura = }  12 \text { years } 3 \text { months }

\text {Lisa = }  11 \text { years } 7 \text { months }

\text {Deborah = }  11 \dfrac{1}{2}  \text { years } =  11 \text { years } 6 \text { months }

\text {Elizabeth = }  11 \dfrac{1}{3}  \text { years } =  11 \text { years } 4 \text { months }

-----------------------------------------------------
Find Total :
-----------------------------------------------------
\text {Total = } 12 \text { yr } 3 \text { mth } + 11 \text { yr } 7 \text { mth } + 11 \text { yr} 6 \text { mth} +11 \text { yr } 4 \text { mth}

\text {Total = } 45 \text { years } 20 \text { months }

-----------------------------------------------------
Find Average :
-----------------------------------------------------
\text {Average = } (45 \text { years } 20 \text { months }) \div 4

\text {Average = } 560 \text { months } \div 4

\text {Average = } 140 \text { months } \div 4

\text {Average = } 11 \text { years } 8 \text { months }

----------------------------------------------------------------------------------------------------------
\bf \text {Answer : Average = } 11 \text { years } 8 \text { months }
----------------------------------------------------------------------------------------------------------
6 0
4 years ago
Solve pls brainliest! the first one to answer it gets the brainliest
mart [117]

Answer:

2.4

Step-by-step explanation:

5 0
3 years ago
I'll give a brainliest <br>Find the integers which satisfy the inequality. - 5 &lt; 2n - 1 ≤ 5<br>​
ella [17]

Answer:

-1, 0, 1, 2, 3

Step-by-step explanation:

Solve two equations:

-5 < 2n-1 => -4 < 2n => n > -2

and

2n-1 ≤ 5 => 2n ≤ 6 => n ≤ 3

so

-2 < n ≤ 3

then enumerate the possible values for n

-1, 0, 1, 2, 3

4 0
3 years ago
Read 2 more answers
An externally imposed project completion time sets the project duration at 75 days. The critical path of the project is establis
Katen [24]

Answer:

3 days

Step-by-step explanation:

This can be seen under Schedule management (Initiating)

Project float is simply the difference between the completion period externally imposed and the estimated time duration of the project (usually this is equal to the critical path length). The difference in this case is 0 because

Externally imposed project date = 75 days

Estimated duration of project = 75 days

Hence, a difference in these will be 0

However, a standard deviation of 3 days was giving (This implies that there is a +/- 3 days in the critical path from the estimated project duration) .

Hence the maximum float in this case will be 0 + 3 = 3 days

8 0
4 years ago
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