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Oxana [17]
3 years ago
11

How is the time, in hours, related to the number of times 2 is used as a factor

Mathematics
1 answer:
nata0808 [166]3 years ago
4 0
What do you mean because i dont understand the wording u use


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Determine whether lines L1 and L2 passing through the pairs of points are parallel, perpendicular, or neither. L1 : (–5, –5), (4
Mazyrski [523]

Answer:  The lines L1 and L2 are parallel.

Step-by-step explanation:  We are given to determine whether the following lines L1 and L2 passing through the pair of points are parallel, perpendicular or neither :

L1 : (–5, –5), (4, 6),

L2 : (–9, 8), (–18, –3).

We know that a pair of lines are

(i) PARALLEL if the slopes of both the lines are equal.

(II) PERPENDICULAR if the product of the slopes of the lines is -1.

The SLOPE of a straight line passing through the points (a, b) and (c, d) is given by

m=\dfrac{d-b}{c-a}.

So, the slope of line L1 is

m_1=\dfrac{6-(-5)}{4-(-5)}=\dfrac{6+5}{4+5}=\dfrac{11}{9}

and

the slope of line L2 is

m_2=\dfrac{-3-8}{-18-(-9)}=\dfrac{-11}{-9}=\dfrac{11}{9}.

Therefore, we get

m_1=m_2\\\\\Rightarrow \textup{Slope of line L1}=\textup{Slope of line L2}.

Hence, the lines L1 and L2 are parallel.

5 0
3 years ago
If the volume of the pyramid shown is 108 inches cubed, what is the area of its base?
Yuri [45]

Answer: 81/2 inches squared or 40.5 inches squared

Step-by-step explanation:

The volume of a pyramid = bh/3

so V = bh/3

V = 108 and h = 8, so plug these in

108 = 8b/3

b = 108 *3/8

= 40.5 inches squared

or if you want a fraction, 81/2 inches squared

7 0
3 years ago
What is the area of that figure??
pochemuha

Answer:

It's 51 square inches. You can multiply 12x8 and subtract 9x5 from that.

4 0
3 years ago
Use a transformation to solve the equation. w/4 = 8 can you also leave a detailed explanation on how this equation = 32
OverLord2011 [107]

Answer:

w=32

Step-by-step explanation:

w/4 = 8

4/1 * w/4 = 8*4

w=32

5 0
3 years ago
Find the limit of the function algebraically.
ArbitrLikvidat [17]
\bf \lim\limits_{x\to -10}~\cfrac{x^2-100}{x+10}\implies \cfrac{\stackrel{\textit{difference of squares}}{x^2-10^2}}{x+10}\implies \cfrac{(x-10)(\underline{x+10})}{\underline{x+10}}
\\\\\\
\lim\limits_{x\to -10}~x-10\implies \lim\limits_{x\to -10}~-10-10\implies -20
5 0
3 years ago
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