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3 years ago

Plz help me out !! Am rlly bad at these type of math

Mathematics

2 answers:

Sever21 [200]3 years ago

8 0

Answer:

Ok look times 18 and 10 then what ever you get with that times it with 12

Step-by-step explanation:

kakasveta [241]3 years ago

6 0

Answer:128

Step-by-step explanation:

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$\bf \displaystyle \cfrac{d}{dx}\left[ \int\limits_{\sqrt{x}}^{\frac{\pi }{2}}~[cos(t^2)]dt \right]\implies \cfrac{d}{dx}\left[ -\int\limits_{\frac{\pi }{2}}^{\sqrt{x}}~[cos(t^2)]dt \right]\\\\\\ \cfrac{d}{dx}\left[ -\int\limits_{\frac{\pi }{2}}^{x^{\frac{1}{2}}}~[cos(t^2)]dt \right]\\\\
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6 0

3 years ago

There is a 15% discount on all items in a store. You buy a $12 shirt and $45 jeans. What will be your total cost?

Nina [5.8K]

Answer:

The total cost of the shirt and Jean's together with a 15% discount would be $48.45.

Step-by-step explanation:

if you at 12 and 45 you get 57 now if you multiply that by .15 which would be 15% you would get $48.45.

6 0

3 years ago

Read 2 more answers

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