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Daniel [21]
2 years ago
13

Help please I’ll make it brainliest!!!!!

Mathematics
1 answer:
mr Goodwill [35]2 years ago
4 0

Step-by-step explanation:

it is very simple, once you remember that "kilo" means 1000, "mili" means 1/1000, and "centi" means 1/100.

and therefore 1 cg = 10 mg, or 1 cm = 10 mm.

"deci" means 1/10.

and therefore 1 dm = 100 mm, 1dg = 100 mg.

1.

9.32 kg = 9.32×1000 = 9320 g

2.

1.429 g = 1.429/1000 = 0.001429 kg

3.

287 g = 287/1000 = 0.287 kg

4.

4.6 L = 4.6×1000 = 4600 mL

5.

0.119 L = 0.119×1000 = 119 mL

6.

9936 mL = 9936/1000 = 9.936 L

7.

26793 mL = 26793/1000 = 26.793 L

8.

0.06 L = 0.06×1000 = 60 mL

9.

170 cg = 170×10 = 1700 mg

10.

2674 cm = 2674/100 = 26.74 m

11.

9.05 mm = 9.05/100 = 0.0905 dm

12.

2 L = 2×1000 = 2000 mL

13.

62.4 L = 62.4×1000 = 62400 mL

14.

99.9 mm = 99.9/1000 = 0.0999 m

15.

4.34 g = 4.34×100 = 434 cg

16.

10 km = 10×1000 m = 10×1000×1000 = 10000000 mm

17.

65 cL = 65/100 = 0.65 L

18.

105 mL = 105/1000 = 0.105 L

19.

0.27 g = 0.27×100 = 27 cg

20.

7777 m = 7777/1000 = 7.777 km

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To find the intercepts of this give function of g(n), we have to find both the points present on the axis. That is, X-Intercept axial or axis point and the Y-Intercept axial or axis point and apply the zero factor principle to get the actual points on the graph for both the respective intercepts. Let me make it simpler, by showing the whole process via the LaTeX interpreter equation editor.

The X-Intercept is that actual point present in the graphical interpretation where the Y-axis is taken as zero, this makes us to point out the position of X-Intercept points on its X-axis and Y-axis. Take the variable "n" as the variable of "x", it will not change any context or such, we can take any variables for calculations, it does not hinder the processing of Intercepts for the axial points on a graph.

\boxed{\mathbf{\therefore \quad -2(3x - 1)(2x + 1) = 0}}

By the zero factor principle, both of them can be separately calculated as a zero on their either sides of the expression.

\mathbf{\therefore \quad 3x - 1 = 0}

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\mathbf{\therefore \quad x = \dfrac{1}{3}}

Similarly, for the second X-Intercept point for the value of 0 in the Y-axis or Y axial plane in a 2 dimensional Graphical representation is going to be, As per the zero factor principle:

\mathbf{\therefore \quad 2x + 1 = 0}

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\boxed{\mathbf{\underline{X-Intercept: \quad \Bigg(\dfrac{1}{3}, \: 0 \Bigg), \: \: \: \: \Bigg(-\dfrac{1}{2}, \: 0 \Bigg)}}}

Therefore, for our Y-Intercept axial point the X axial plane will instead turn out to be a value with zero on a Graphical representation to obtain the actual points for Y-axis and the Y-Intercept for x = 0 as a point on the graph itself.

Just substitute the value of "0" in "x" axis as a variable on the provided expression. Therefore:

\boxed{\mathbf{= -2(3 \times 0 - 1) (2 \times 0 + 1)}}

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\mathbf{y = - 2 (- 1) (0 + 1)}

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