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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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A. 0.333<br> B. 6<br> C. 0.667<br> D. 3
kow [346]

c

Answer:

Step-by-step explanation:

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3 years ago
An element with mass 590 grams decays by 19. 5% per minute. How much of the element is remaining after 15 minutes, to the neares
solmaris [256]

Answer:

Step 1

Formulate a recursive sequence modeling the number of grams after n  minutes.

we have that

100%-17.1%-------------- > 82.9%------------> 0.829

a(n) = 780*[0.829^n]

for n=19 minutes

a(19)=780*[0.829^(19)]=22.1121 g---------------> 22.1 g

the answer is 22.1 g

Step-by-step explanation:

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2 years ago
which of the following lengths of lines would be used to create a right triangle 6,9,12 5,12,13 4,8,16 7,15,19.7
marin [14]
If this is talking about the Pythagorean theorem then it would be 5,12,13 becuase because the equation for the triangle is
2 2 2
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6 0
3 years ago
6= a/4 +2<br><br>two step equations ​
Marianna [84]

Answer: A=16

Step-by-step explanation:

24=a+8

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4 0
3 years ago
Read 2 more answers
[(1/(3+x))-(1/3)] / x<br> Limit 0
Effectus [21]
First find a common denominator and combine the fractions in the numerator:

\displaystyle\lim_{x\to0}\frac{\dfrac1{3+x}-\dfrac13}x=\lim_{x\to0}\frac{\dfrac3{3(3+x)}-\dfrac{3+x}{3(3+x)}}x=\lim_{x\to0}\frac{3-(3+x)}{3x(3+x)}

Now simplify and cancel out all the terms that you can:

\displaystyle\lim_{x\to0}\frac{3-3-x}{3x(3+x)}=-\frac13\lim_{x\to0}\frac1{3+x}

Since the remaining expression is continuous as a function of x, you can directly substitute to end up with

\displaystyle-\frac13\lim_{x\to0}\frac1{3+x}=-\frac13\times\frac1{3+0}=-\frac19
8 0
3 years ago
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