Describe how to draw a number line and graph –5 and –(–5) on it.

Mathematics

2 answers:

Sloan [31]4 years ago

8 0

Answer:

see below

Step-by-step explanation:

Draw a line with points on it from -10 to 10 going by 1

Put a dot at -5 for the -5 point

-(-5) is 5

Put a dot at 5

almond37 [142]4 years ago

3 0

Answer:

Draw a straight horizontal line, and mark integers on it in increasing order as you move towards the right

Start counting from 0.

For -5, place a mark 5 units towards left of 0

For -(-5), which is actually +5, place a mark 5 units towards right of 0

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<img src="https://tex.z-dn.net/?f=%20-%2032%20%2B%204%20x%20%3D%205%28%20-%207%20%2B%20x%29" id="TexFormula1" title=" - 32 + 4 x

harkovskaia [24]

$\bf -32+4x=5(-7+x)\implies -32+4x=\stackrel{\textit{distributing}}{-35+5x} \\\\\\ \stackrel{\textit{subtracting 4x to both sides}}{-32=-35+x}\implies \stackrel{\textit{adding 35 to both sides}}{3=x}$

7 0

4 years ago

What is 80%of 40equals what

kari74 [83]

If you would like to solve 80% of 40, you can do this using the following steps:

80% of 40 = 80% * 40 = 80/100 * 40 = 32

The correct result would be 32.

7 0

3 years ago

Read 2 more answers

The following statement is either true (in all cases) or false (for at least one example). If false, construct a specific e

matrenka [14]

Answer:

False, counterexample below

Step-by-step explanation:

Denote the unit vectors of R^3 by $e_1=(1,0,0), e_2=(0,1,0), e_3=(0,0,1)$ . Now consider $v_1=e_1, v_2=2e_1$ and $v_3=e_3$ . We have that $v_1, v_2,v_3 \in \mathbb{R}^3$ . Also, the vector $v_3$ is not a linear combination of $v_1, v_2$ because any linear combination of these two vectors will have third coordinate zero, but v_3 has third coordinate 1 so they can't be equal.

However, the set $\{v_1, v_2,v_3\}$ is not linearly independent, because $2v_1-v_2+0v_3=0$ is a non-trivial linear combination of these vectors that equals zero.