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nekit [7.7K]
3 years ago
9

How many reasons must your thesis contain and which reason is first

Mathematics
1 answer:
finlep [7]3 years ago
5 0
<span><span>The more sources the better
Every researcher aims to have a standardized and uniform study when a study will be replicated or repeated for various reasons.

</span>The number of accounts and evidential support of your thesis must be consistent and be sufficient and succinct to the topic of study. These reasons and evidential support should claim proximate with the pronounced experimentation and variables which will be studied. In sense, the more accounts and sources the better is your study, the greater it will stand out and present itself. This is important for every thesis paper or dissertation since these accounts provides and explicitly discusses the studies that have been undertaken before which can greatly influence the current study's credibility and reliability. </span>

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Does anyone know?...
Inessa [10]

Answer:

NOPE

Step-by-step explanation:

6 0
3 years ago
Find three rational numbers between (-4) and (-2)
zavuch27 [327]
-1/8 -2/8 -3/8 because any fraction is ok, they don't have 0 as a denominator.
4 0
3 years ago
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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
What is the coordinate of the midpoint of segment CE?​
Anna35 [415]

Answer:

-1 :) you can count the number of spaces, divide by 2, and count to the middle by that ammount ;)

7 0
3 years ago
What is the solution to -6z=72 1) 12 2) 9 3) -9 4) -12
S_A_V [24]

Step-by-step explanation:

-6z = 72

z = 72/(-6)

= -12

4) -12

6 0
4 years ago
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