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KIM [24]
2 years ago
7

I will mark you as brainliest if you're answer is correct​

Mathematics
1 answer:
GalinKa [24]2 years ago
8 0

Answer:

perimeter = 28.13 cm

Step-by-step explanation:

Calculate the length of a line drawn from A to C.

sin 85 = b/8, b = 7.97

cos 85 = c/8, c = 0.697

AC² =7.97² + (6 + 0.697)² = 63.52 + 44.85 = 108.37

AC = 10.41

Now you have a right triangle ADC with a known hypotenuse and  one leg.

Using the Pythagorean theorem:

DC² = 10.41² - 5²

DC² = 108.37 - 25 = 83.37

DC = 9.13 cm

perimeter = 5 + 6 + 8 + 9.13 = 28.13 cm

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The hypotenuse and one of the legs of a right triangle form an angle that has a cosine of 2 2 . what is the measure of the angle
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The correct question is
The hypotenuse and one of the legs of a right triangle form an angle that has a cosine of √<span>2/2 . 
What is the measure of the angle? 

Let
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[distance of one of the leg/hypotenuse]=√2/2

<span>I could say that
</span>distance of one of the leg=√2
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so
<span>applying the Pythagorean theorem
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3 years ago
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OlgaM077 [116]

so, let's keep in mind that

\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}

so let's make a quick table of those solutions, say A, B, C solutions with x,y,z liters of acid, with an acidity of 0.25, 0.40 and 0.60 respectively.


\bf \begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ A&x&0.25&0.25x\\ B&y&0.40&0.4y\\ C&z&0.60&0.6z\\ \cline{2-4}&\\ mixture&78&0.45&35.1 \end{array} \\\\\\ \begin{cases} x+y+z=78\\ 0.25x+0.4y+0.6z=35.1 \end{cases}


we know she's using "z" liters and those are 3 times as much as "y" liters, so z = 3y.


\bf \begin{cases} x+y+3y=78\\ x+4y=78\\[-0.5em] \hrulefill\\ 0.25x+0.4y+0.6(3y)=35.1\\ 0.25x+0.4y=1.8y=35.1\\ 0.25x+2.2y=35.1 \end{cases}\implies \begin{cases} x+4y=78\\\\ 0.25x+2.2y=35.1 \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ x+4y=78\implies \boxed{x}=78-4y \\\\\\ \stackrel{\textit{using substitution on the 2nd equation}}{0.25\left( \boxed{78-4y} \right)+2.2y=35.1}\implies 19.5-y+2.2y=35.1


\bf 1.2y=15.6\implies y=\cfrac{15.6}{1.2}\implies \blacktriangleright y=13 \blacktriangleleft \\\\\\ x=78-4y\implies x=78-4(13)\implies \blacktriangleright x=26 \blacktriangleleft \\\\\\ z=3y\implies z=3(13)\implies \blacktriangleright z=39 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{25\%}{26}\qquad \stackrel{40\%}{13}\qquad \stackrel{60\%}{39}~\hfill

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