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miv72 [106K]
3 years ago
13

Victoria uses

Mathematics
2 answers:
Gennadij [26K]3 years ago
6 0

Answer:

B.10 2/3

Step-by-step explanation:

edgems 2020-2021

sineoko [7]3 years ago
3 0

Answer:

B. or 10 2/3

Step-by-step explanation: Because if she had only 2 cups left then 1 cup can make 5 drinks so 10 then the rest would be a fraction.

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After increasing the price of an item by 20% the price was $300.00. What was the original price?​
mr_godi [17]

the answer to this question is $250

4 0
3 years ago
What's |5| + |–7|?
Marizza181 [45]

Answer:

C) 12

Step-by-step explanation:

Absolute value always means it is the positive version of the number

This switches -7 to 7 and 5 stays the same.

5 + 7 = 12

4 0
3 years ago
Read 2 more answers
If f(x) =4x^2 + 1 and g(x) =x2 -5, find (f+g)(x)
Lilit [14]

The value of (f+g)(x) is 5 x^{2}-4

Explanation:

Given that the functions f(x)=4 x^{2}+1 and g(x)=x^{2}-5

We need to determine the value of (f+g)(x)

The value of (f+g)(x) can be determined by substituting the value of f(x) and g(x) and simplifying the terms.

Thus, let us assign f(x)=4 x^{2}+1 in the function (f+g)(x), we have,

4 x^{2}+1+g(x)

Now, let us assign g(x)=x^{2}-5 in the function (f+g)(x), we get,

4 x^{2}+1+x^{2}-5

Grouping the like terms, we have,

4 x^{2}+x^{2}+1-5

Adding the like terms, we get,

5 x^{2}-4

Hence, the value of (f+g)(x) is 5 x^{2}-4

3 0
3 years ago
Roger can run one mile in 8 minutes. jeff can run one mile in 6 minutes. if jeff gives roger a 1 minute head​ start, how long wi
Anastasy [175]
Recall your d = rt, distance = rate * time.

now, if Roger can do 1 mile in 8 minutes, so in 1 minute, he has done then 1/8 of a mile, so his rate is 1/8 miles per minute.

if Jeff can do 1 mile in 6 minutes, he's faster, in 1 minute he has done 1/6 of a mile, so his rate is 1/6 miles per minute.

now, when Jeff catches up with Roger, the distance covered by both will be the same, say "d" miles, because, at that millisecond, Jeff will be neck and neck with Roger, and their covered distance will be the same.

now, Jeff is generous and let Roger roll on for 1 minute before him, so, by the time time Roger has covered "d" miles, he has been running for say "t" minutes.

however, since Jeff started later by 1 minute, he hasn't been running for "t" minutes, but for "t - 1" minutes.

\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mpm}{rate}&\stackrel{minutes}{time}\\
&------&------&------\\
Roger&d&\frac{1}{8}&t\\\\
Jeff&d&\frac{1}{6}&t-1
\end{array}
\\\\\\
\begin{cases}
\boxed{d}=\frac{1}{8}t\\\\
d=\frac{1}{6}(t-1)\\
------\\
\boxed{\frac{1}{8}t}=\cfrac{t-1}{6}
\end{cases}
\\\\\\
\cfrac{t}{8}=\cfrac{t-1}{6}\implies 6t=8t-8\implies 8=2t\implies \cfrac{8}{2}=t\implies \boxed{\stackrel{mins}{4}=t}
4 0
3 years ago
What property is being shown? <br><br> 5x^3 * 1 = 5x^3
maw [93]

Let's call that the Multiplicative Identity Property.

1 is the identity element for multiplication.  Those are just fancy words for the obvious property that anything times 1 is itself.


4 0
3 years ago
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