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Setler [38]
3 years ago
8

What is the GCF of 33,55,132

Mathematics
1 answer:
valina [46]3 years ago
7 0
Greatest\ common\ factor\ is\ the\ highest\ number\ by\ witch\ 33\ and\ 55\\and\ 132\ can\ be\ divided\\\\
33:3\\
11:11\\\\
55:5\\
11:11\\\\
132:2\\66:2\\33:3\\11:11
\\\\GCF=3*11=33\\\\Greatest\ common\ factor\ is\ equal\ to\ 33.

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The difference of d minus 3 divided by 2
Eduardwww [97]

Answer:

(d-3)/2

Step-by-step explanation:

Difference is subtraction

(d-3)/2

6 0
3 years ago
A bicycle shop sells only bicycles and tricycles. Altogther there are 23 seats and 50 wheels. How many bicycles and tricycles ar
slamgirl [31]

Answer:

The number of bicycles is 19.

and the number of tricycles is 4.

Step-by-step explanation:

As we know that the number of seats in bicycles as well as in tricycles is 1.

The number of wheels in bicycles is 2

and the number of wheels in tricycles is 3.

Let the number of bicycles be x.

and the number of tricycles be y.

Thus using the given information we can make equation as,

x + y = 23

and, 2x + 3y = 50

Solving these two equations:

We get, x = 19 and y = 4

Thus the number of bicycles is 19.

and the number of tricycles is 4.

8 0
3 years ago
In a class of 40 students, 30 read
DIA [1.3K]
Ok so 24 of them and that would be ur answer
6 0
3 years ago
What is the simplified form of 400x100
erma4kov [3.2K]

Answer:

4x1

Step-by-step explanation:

5 0
3 years ago
Geometric Series assistance
Levart [38]

we have been asked to find the sum of the series

\sum _{n=1}^5\left(\frac{1}{3}\right)^{n-1}

As we know that a geometric series has a constant ratio "r" and it is defined as

r=\frac{a_{n+1}}{a_n}=\frac{\left(\frac{1}{3}\right)^{\left(n+1\right)-1}}{\left(\frac{1}{3}\right)^{n-1}}=\frac{1}{3}

The first term of the series is a_1=\left(\frac{1}{3}\right)^{1-1}=1

Geometric series sum formula is

S_n=a_1\frac{1-r^n}{1-r}

Plugin the values we get

S_5=1\cdot \frac{1-\left(\frac{1}{3}\right)^5}{1-\frac{1}{3}}

On simplification we get

S_5=\frac{121}{81}

Hence the sum of the given series is \frac{121}{81}

5 0
3 years ago
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