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3 years ago

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Fred is making a bouquet of carnationsand roses. The carnations cost $5.25 in all. The roses are $1.68 each. If Fred spent $18.6

9 all together, how many roses did he use? I will brainliest
a.5 b.6 c.7 d.8

2 answers:

Natalija [7]3 years ago

8 0

Answer:

I think its d.8. it sounds like the correct answer.

kotegsom [21]3 years ago

4 0

Answer:

8 roses

Step-by-step explanation:

18.69-5.25=13.44

13.44÷1.68=8 Hope this helps.

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Will mark brainlest how to multipy it

Lesechka [4]

Answer:

$\displaystyle x = 2 \sqrt{3}$

Step-by-step explanation:

we would like to solve the following equation:

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in order to do so do cross multiplication:

$\displaystyle x \sqrt{3} = 6$

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since we ended up with a square root on the denominator so we can consider rationalising the denominator to do so multiply both numerator and denominator by √3 which yields:

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Ok, now i need help from 1-5, 20 points

ratelena [41]

1 = 8

2=3+8

3=12

4=9

8 that's it bra

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VikaD [51]

Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

$\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
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b=opposite\\
\end{cases}
\\\\\\
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recall that

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
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csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
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sec(\theta)=\cfrac{hypotenuse}{adjacent}$

therefore, let's just plug that on the remaining ones,

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\\\\\\
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tan(\theta)=\cfrac{-1}{\sqrt{35}}
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cot(\theta)=\cfrac{\sqrt{35}}{1}
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now, let's rationalize the denominator on tangent and secant,

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\\\\\\
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