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2 years ago

11

For tanisha's lemonade recipe, 14 lemons are required to make 7 cups of lemonade. at what rate are lemons being used in lemons p

er cup of lemonade?

1 answer:

Aneli [31]2 years ago

7 0

I think it’s 2 because 14 divided 7 is 2

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2 years ago

Find the value of each expression using the given information.

Slav-nsk [51]

Recall that $\cos^2\theta+\sin^2\theta=1$ . So

$\sin\theta=\pm\sqrt{1-\cos^2\theta}$

Given that both $\cos\theta$ and $\tan\theta$ , and knowing that $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$ , it follows that we should expect $\sin\theta>0$ , so we take the positive root above.

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3 years ago

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3 years ago

Match each value with its formula for ABC.

MariettaO [177]

The solution to the question is:

c is 6 = $\sqrt{a^{2} + b^{2} -2abcosC }$

b is 5 = $\sqrt{a^{2} + c^{2} -2accosB }$

cosB is 2 = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

a is 4 = $\sqrt{b^{2} + c^{2} -2bccosA }$

cosA is 3 = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

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<h3>What is cosine rule?</h3>

it is used to relate the three sides of a triangle with the angle facing one of its sides.

The square of the side facing the included angle is equal to the some of the squares of the other sides and the product of twice the other two sides and the cosine of the included angle.

Analysis:

If c is the side facing the included angle C, then

$c^{2}$ = $a^{2}$ + $b^{2}$ -2ab cos C-----------------1

then c = $\sqrt{a^{2} + b^{2} -2abcosC }$

if b is the side facing the included angle B, then

$b^{2}$ = $a^{2}$ + $c^{2}$ -2accosB-----------------2

b = $\sqrt{a^{2} + c^{2} -2accosB }$

from equation 2, make cosB the subject of equation

2ac cosB = $a^{2}$ + $c^{2}$ - $b^{2}$

cosB = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

if a is the side facing the included angle A, then

$a^{2}$ = $b^{2}$ + $c^{2}$ -2bccosA--------------------3

a = $\sqrt{b^{2} + c^{2} -2bccosA }$

from equation 3, making cosA subject of the equation

2bcosA = $b^{2}$ + $c^{2}$ - $a^{2}$

cosA = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

from equation 1, making cos C the subject

2abcosC = $b^{2}$ + $a^{2}$ - $c^{2}$

cos C = $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

In conclusion,

c is 6 = $\sqrt{a^{2} + b^{2} -2abcosC }$

b is 5 = $\sqrt{a^{2} + c^{2} -2accosB }$

cosB is 2 = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

a is 4 = $\sqrt{b^{2} + c^{2} -2bccosA }$

cosA is 3 = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

cosC is 1 = $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

Learn more about cosine rule: brainly.com/question/4372174

$SPJ1

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2 years ago

C ( , )D ( , ) E ( , ) F ( , )

Alex777 [14]

As shown in the question, the translation changes the points as follows:

$(x,y)\rightarrow(x+2,y-2)$

Therefore:

$\begin{gathered} C(-5,-1)\rightarrow C^{\prime}(-3,-3) \\ D(-4,3)\rightarrow D^{\prime}(-2,1) \\ E(-3,2)\rightarrow E^{\prime}(-1,0) \\ F(-2,0)\rightarrow F^{\prime}(0,-2) \end{gathered}$

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1 year ago