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worty [1.4K]
3 years ago
5

How to figure out witch rays are parallel

Mathematics
1 answer:
leonid [27]3 years ago
7 0
Parallel lines are the lines that are always the same distance and never touch.
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Question 1b? please help?
Alborosie
I hope this may help.u

5 0
3 years ago
I need help with this??
Marina86 [1]

Select both the 2nd and 4th options!

3 0
3 years ago
Someone please HELP!?
jeka94
 A negative exponent just means that the base is on the wrong side of the fraction line, so you need to flip the base to the other side.

For example

x^{-2}=\frac{1}{x^2}

or

( \frac{2}{5} )^{-4}=( \frac{5}{2} )^4

***************************************
(8r^{-5})^{-3}

(8* \frac{1}{r^5} )^{-3} \\\\( \frac{8}{r^5} )^{-3}
 \\  \\ ( \frac{r^5}{8} )^3
\\\\ \mathrm{Apply\:exponent\:rule}:\quad *\left(\frac{a}{b}\right)^c=\frac{a^c}{b^c} \ \ \ \ \ \ \ \ \ \ \ \   *(a^b)^c=a^{bc}
 \\\\( \frac{r^{15}}{8^3} )
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The answer is "D"


7 0
3 years ago
Steel rods are manufactured with a mean length of 29 centimeter (cm). Because of variability in the manufacturing process, the l
zavuch27 [327]

Solution :

Given data :

The mean length of the steel rod = 29 centimeter (cm)

The standard deviation of a normally distributed lengths of rods = 0.07 centimeter (cm)

a). We are required to find the proportion of rod that have a length of less than 28.9 centimeter (cm).

Therefore, P(x < 28.9) = P(z < (28.9-29) / 0.07)

                                    = P(z < -1.42)

                                   = 0.0778

b). Any rods which is shorter than 24.84 cm or longer than 25.16 cm that re discarded.

Therefore,

P (x < 24.84 or 25.16 < x) = P( -59.42 < z or -54.85)

                                         = 1.052

4 0
3 years ago
What is the missing reason for line 5 in this proof?
Molodets [167]
Line 5 shows the division property of equality. 

The division property of equality states that you can divide both sides of the equation by the same number and the equation remains the same. This equation is being divided by 4 on both sides.
5 0
4 years ago
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