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ozzi
3 years ago
8

Can a pair of lines be both parallel and perpendicular? Explan

Mathematics
2 answers:
nasty-shy [4]3 years ago
8 0
No because for two lines to be parallel they can not intersect each other but a different line can.
weeeeeb [17]3 years ago
5 0
No, the 2 lines can never ever be both parallel and perpendicular if I'm not mistaken. This is because a set of parallel lines will never touch each other at all. However perpendicular lines are two lines that meet up to get an angle of 90. You cannot have two lines that never touch and touch at the same time.

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A town's population is growing at a steady rate of 6% per year, so it's population after t years is given by P = 5.26(1.06)t, wh
nydimaria [60]
Hi,
I want to answer yr question, that is our earth. Is that right.
7 0
3 years ago
Surface area of cuboid 20 mm 8mm 11mm
SpyIntel [72]

Answer:

936mm^2

Step-by-step explanation:

Surface area of cuboid 20 mm 8mm 11mm

Surface area of the cuboid=2(20×8+8×11+11×20)

=2(160+88+×220)

=2(468)

=936mm^2

So the answer is936mm^2

7 0
3 years ago
I need the answer please and can u answer my recent to.
ale4655 [162]

Answer:

30 hours.

Step-by-step explanation:

Richard can build 15 snowballs in 1 hour

But 2 snowballs melt every 15 minutes

In 1 hour, the number of snowballs that will have melted is:

\frac{60}{15} × 2 = 8

The number of snowballs that will have remained = 15 - 8 = 7

So 7 snowballs will have remained in 1 hour

For 210 snowballs to have remained, it will take:

\frac{210}{7} × 1 = 30 hours.

3 0
3 years ago
What is the place of 9,201,779 and 2,008,277
swat32
The places are Tens and Millions 

4 0
4 years ago
Houng deposits $2,000 into an account that earns 4.2% interest compounded daily. What is the annual percentage yield to the near
jek_recluse [69]
Assuming a year has 365 days, a daily compounding interest is equivalent to having a compounding cycle of 365 per year, thus

\bf ~~~~~~  \textit{Annual Yield Formula}
\\\\
~~~~~~~~~~~~ \left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 4.2\%\to \frac{4.2}{100}\to &0.042\\
n=
\begin{array}{llll}
\textit{times it compounds per year}
\end{array}\to &365
\end{cases}
\\\\\\
 \left(1+\frac{0.042}{365}\right)^{365}-1\qquad \approx \qquad  0.042891958856893
\\\\\\
0.042891958856893\cdot 100\qquad \approx \qquad \stackrel{\%}{4.289}
7 0
3 years ago
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