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SashulF [63]
3 years ago
12

Can someone explain how to solve this?

Mathematics
1 answer:
White raven [17]3 years ago
6 0
An incenter is equidistant to the sides of the triangle so PN = PO and PN = PM. PO = 21; LO = 27; PL = 42
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10. Weiming and Joyce intend to buy a present for Siti. Weiming agrees to pay at least twice as much as but at most $150 more th
djverab [1.8K]

The  required amount Weiming has to pay is given by the limit of $150

more than the amount Joyce has to pay.

Response:

  • The greatest amount Weiming has to pay is<u> $180</u>

<h3>Which method can be used to analyze the amount paid by Weiming and Joyce?</h3>

Amount Weiming agrees to pay ≥ 2 × Joyce agrees to pay

Amount Weiming agrees to pay ≤ 150 + Amount Joyce agrees to pay

Let <em>x</em> represent the amount Weiming agrees to pay, and let <em>y</em> represent

the amount Joyce agrees to pay, we have;

x + y = 210

x ≥ 2·y

x ≤ y + 150

When the amount Weiming is $150 more than Joyce, which is the

limiting amount for Weiming, we have;

x = y + 150

Which gives;

y + 150 + y = 210

2·y = 210 - 150 = 60

y = 60 ÷ 2 = 30

x = 30 + 150 = 180

Therefore;

  • The greatest amount Weiming has to pay is, x = <u> $180</u>

Learn more about inequalities here:

brainly.com/question/17882496

4 0
3 years ago
Read 2 more answers
Linear e
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5 0
3 years ago
√32 greater or less than 5.1
castortr0y [4]
The pie of 32 is 5.6 and then there's 5.1

Your answer will be greater because 5.6 is greater than 5.1.
7 0
4 years ago
If z1= 3+3i and z2=7(cos(5pi/9) + i sin (5pi/9)), then z1/z2= blank
mixas84 [53]

z1=\stackrel{a}{3}+\stackrel{b}{3}i~~ \begin{cases} r = \sqrt{a^2+b^2}\\ r = \sqrt{18}\\[-0.5em] \hrulefill\\ \theta =\tan^{-1}\left( \frac{b}{a} \right)\\ \theta =\frac{\pi }{4} \end{cases}~\hfill z1=\sqrt{18}\left[\cos\left( \frac{\pi }{4} \right) i\sin\left( \frac{\pi }{4} \right) \right] \\\\[-0.35em] ~\dotfill

\cfrac{z1}{z2}\implies \cfrac{\sqrt{18}\left[\cos\left( \frac{\pi }{4} \right) i\sin\left( \frac{\pi }{4} \right) \right]} {7\left[\cos\left( \frac{5\pi }{9} \right) i\sin\left( \frac{5\pi }{9} \right) \right]} \\\\[-0.35em] ~\dotfill\\\\ \qquad \textit{division of two complex numbers} \\\\ \cfrac{r_1[\cos(\alpha)+i\sin(\alpha)]}{r_2[\cos(\beta)+i\sin(\beta)]}\implies \cfrac{r_1}{r_2}[\cos(\alpha - \beta)+i\sin(\alpha - \beta)] \\\\[-0.35em] ~\dotfill

\cfrac{z1}{z2}\implies \cfrac{\sqrt{18}}{7}\left[\cos\left( \frac{\pi }{4}-\frac{5\pi }{9} \right)+i\sin\left( \frac{\pi }{4}-\frac{5\pi }{9} \right) \right] \\\\\\ \cfrac{\sqrt{18}}{7}\left[\cos\left( \frac{-11\pi }{36} \right) +i\sin\left( \frac{-11\pi }{36} \right) \right]\implies \cfrac{\sqrt{18}}{7}\left[\cos\left( \frac{83\pi }{36} \right) +i\sin\left( \frac{83\pi }{36} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{z1}{z2}\approx 0.348~~ + ~~0.496i~\hfill

6 0
3 years ago
You only have to solve one of them but please explain how to do it because I have many more
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85. Relative mins are at x=-1 and x=3; relative max x=1
As x > -infinity, y > infinity
As x > infinity, y > infinity
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3 years ago
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