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ziro4ka [17]
3 years ago
14

A bag of marbles has 5 blue marbles, 3 red marbles, and 6 yellow marbles. A marble is drawn from the bag at

Mathematics
2 answers:
alina1380 [7]3 years ago
5 0

Answer:

14

Step-by-step explanation:

Nady [450]3 years ago
5 0

Answer:

14

Step-by-step explanation:

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What is the first four terms in the multiplication or ,geometric , pattern in which the first term is 2 and then each term is mu
kirza4 [7]

Answer:

Sequence will be 2, 14, 98, 686.

Step-by-step explanation:

Since we know in an geometric sequence the terms are in the form are

T_{n}=a(r)^{n-1} in which

a = first term

r = common ratio

n = number of term

Now from this expression we can all terms of the sequence

in which a = 2 and common ratio r = 7

The sequence will be

2, 2.(7), 2.(7)^{2},2.(7)^{3}

Or 2, 14, 98, 686

So the answer is 2, 14, 98, 686.

7 0
4 years ago
Let r vary directly with s and inversely with t. Which equation represents this equation? Assume that a is a constant.
andrew11 [14]
\bf \qquad \qquad \textit{combined variation}
\\\\
\begin{array}{llll}
\textit{\underline{y} varies directly with \underline{x}}\\
\textit{and inversely with \underline{z}}
\end{array}\implies y=\cfrac{kx}{z}\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad  variation
\end{array}\\\\
-------------------------------\\\\
\stackrel{\textit{\underline{r} varies directly with \underline{s} and inversely with \underline{t}}}{r=\cfrac{ks}{t}\qquad \textit{ and since a = k}\qquad  r=\cfrac{as}{t}}
7 0
3 years ago
What is the sum of these mixed numbers 4 2/3+ 3 1/9
Mariana [72]
Hello!

First you have to make the denominators the same so the numbers are comparable.

You can multiply the top and bottom number in 2/3 to get 6/9

Now you add

4 6/9 + 3 1/9 = 7 7/9

The answer is 7 7/9

Hope this helps!
3 0
3 years ago
Read 2 more answers
4/12 + 3/12 = ???????
iris [78.8K]
7/12

..............................
4 0
3 years ago
Read 2 more answers
I need to know if this is correct
vitfil [10]

i thinks its correct.

3 0
3 years ago
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