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atroni [7]
4 years ago
5

Least to greatest 4 1/2, 4.504, 4.43, 113/25

Mathematics
1 answer:
aleksklad [387]4 years ago
8 0
4.43
4 1/2
4.504
113/25
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Classify the following sequence as arithmetic, geometric, neither or both: 2, 6, 18, ...
yan [13]

Answer:

Geometric

Step-by-step explanation:

Does multiplying the same number with each term give the next term?

If so, then it is Geometric Sequence.

Does adding the same number with each term give the next term?

If so, then it is Arithmetic Sequence.

<em>We see that multiplying by 3 gives us each successive term. 2 times 3 is 6 and 6 times 3 is 18. So it is geometric sequence.</em>

<em />

It is NOT arithmetic sequence since we add 4 to 2 to get next number 6, but we have 12 to get next number, which is 18. So not an arithmetic sequence.

Answer is "Geometric"

4 0
3 years ago
Geometry help please ASAP! Double points!
Tresset [83]
I am in geometry now but we haven’t learned that yet
6 0
3 years ago
W - 8 &gt; 5.6 ??????????????????????????????????????
dmitriy555 [2]

Answer:

IF YOU ARE TRIING TO FOND W IT IS 97

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Determine whether (18 + 35) × 4 = 18 + 35 ×4 is true or false. explain
fomenos
The first one  is 212 
and the second one is 158
so yes it is false they do not equal

6 0
3 years ago
The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2
Arada [10]

Given:

Scale factor s=\dfrac{1}{3}

Center of dilation = (4,2)

To find:

The coordinates of the points C' and A.

Solution:

We know that, if a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

The scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Suppose the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using rule (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Hence, the coordinates of Point C' are C'(2,5).

Let us assume that point A is A(m,n).

Using rule (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

m=12+4

m=16

And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

4 0
3 years ago
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