Stacy invested $3400 at 18%

Mathematics

2 answers:

nata0808 [166]3 years ago

8 0

Answer:

So?

Step-by-step explanation:

ozzi3 years ago

8 0

The answer is 18888 yw

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PLEASE HELP ASAP!!!!!

Wittaler [7]

Answer:

$sec(x)$

Step-by-step explanation:

$cos(x)+cos(x)tan^2(x)$

Multiply cos(x) by 1:

$cos(x)*1+cos(x)tan^2(x)$

Factor out cos(x):

$cos(x)(tan^2(x)+1)$

Apply the pythagorean identity ( $tan^2(x)+1=sec^2(x)$ ):

$cos(x)sec^2(x)$

Rewrite $sec^2(x)$ :

$cos(x)(\frac{1}{cos(x)})^2 \\cos(x)\frac{1}{cos^2(x)}$

Cancel cos(x):

$\frac{1}{cos(x)}$

Rewrite:

$sec(x)$

5 0

4 years ago

The figure shown below is composed of a semicircle and a non-overlapping equilateral triangle and contains a hole that is also c

Sloan [31]

Check the picture below.

since we know the radius of the larger semicircle is 8, thus its diameter is 16, which is the length of one side of the equilateral triangle. We also know the smaller semicircle has a radius of 1/3, and thus a diameter of 2/3, namely the lenght of one side of the small equilateral triangle.

now, if we just can get the area of the larger figure and the area of the smaller one and subtract the smaller from the larger, we'll be in effect making a hole/gap in the larger and what's leftover is the shaded figure.

$\bf \stackrel{\textit{area of a semi-circle}}{A=\cfrac{1}{2}\pi r^2\qquad r=radius}~\hspace{10em}\stackrel{\textit{area of an equilateral triangle}}{A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\stackrel{side's}{length}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\left[ \stackrel{\textit{larger figure}}{\cfrac{1}{2}\pi 8^2~~+~~\cfrac{16^2\sqrt{3}}{4}} \right]\qquad -\qquad \left[ \cfrac{1}{2}\pi \left( \cfrac{1}{3} \right)^2 +\cfrac{\left( \frac{2}{3} \right)^2\sqrt{3}}{4}\right]}$

$\bf \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\frac{4}{9}\sqrt{3}}{4} \right] \\\\\\ \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\sqrt{3}}{9} \right]~~\approx~~ 211.38 - 0.37~~\approx~~ 211.01$