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2 years ago

6

6. A restaurant sold 16 servings of lasagna in one night. This represents 64% of the lasagna prepared for the evening. How many

servings of lasagna were prepared?

1 answer:

Sonbull [250]2 years ago

6 0

Answer:

25 Lasagna were prepared

Step-by-step explanation:

Step one:

given data

we are given that

16 servings is 64% of the total

let the total be x

so

64/100*x= 16

0.64*x= 16

divide both sides by 0.64

x= 16/0.64

x=25 Lasagna were prepared

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It's not difficult to compute the values of $A$ and $B$ directly:

$A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}$
$A=\tan^{-1}(\sin\theta)-\dfrac\pi4$

$B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt$
$B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}$
$B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2$

Let's assume $0$ , so that $|\csc\theta|=\csc\theta$ .

Now,

$\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}$
$\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}$
$\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)$
$\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)$

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in $A$ and $B$ .

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3 years ago

0 A pigeon can fly at a speed of 84 kilometers per hour. How long does it take the pigeon to fly 7 kilometers

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The miles per hour is 84 kilometers per 60 minutes (7:5 ratio) so if we take seven kilometers times the ratio 5/7 we get 5 minutes

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3 years ago

PLEASE I ONLY HAVE 30 MIUNUTES TO FINISH!!!!!!!!!!!!!

creativ13 [48]

Answer:

When it is small & when the sample has things that are different from the population.

Step-by-step explanation:

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3 years ago

Read 2 more answers

A hole is poked in a balloon so that every minute the balloon loses 3/10ft 3 of air. What is the total change in the number of c

GaryK [48]

5/9 minutes

amount lost=times times amout lost per unit of time
amount lost=5/9 times 3/10
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3 years ago

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Veseljchak [2.6K]

cos θ = $\frac{-4\sqrt{65} }{65}$ , sin θ = $\frac{-7\sqrt{65} }{65}$ , cot θ = 4/7, sec θ = $\frac{-\sqrt{65} }{4}$ , cosec θ = $\frac{-\sqrt{65} }{7}$

<h3>What are trigonometric ratios?</h3>

Trigonometric Ratios are values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle.

Sin θ: Opposite Side to θ/Hypotenuse

Tan θ: Opposite Side/Adjacent Side & Sin θ/Cos

Cos θ: Adjacent Side to θ/Hypotenuse

Sec θ: Hypotenuse/Adjacent Side & 1/cos θ

Analysis:

tan θ = opposite/adjacent = 7/4

opposite = 7, adjacent = 4.

we now look for the hypotenuse of the right angled triangle

hypotenuse = $\sqrt{7^{2} + 4^{2} }$ = $\sqrt{49+16}$ = $\sqrt{65}$

sin θ = opposite/ hyp = $\frac{7}{\sqrt{65} }$

Rationalize, $\frac{7}{\sqrt{65} }$ x $\frac{\sqrt{65} }{\sqrt{65} }$ = $\frac{7\sqrt{65} }{65}$

But θ is in the third quadrant(180 - 270) and in the third quadrant only tan and cot are positive others are negative.

Therefore, sin θ = - $\frac{7\sqrt{65} }{65}$

cos θ = adj/hyp = $\frac{4}{\sqrt{65} }$

By rationalizing and knowing that cos θ is negative, cos θ = - $\frac{-4\sqrt{65} }{65}$

cot θ = 1/tan θ = 1/7/4 = 4/7

sec θ = 1/cos θ = 1/ $\frac{4}{\sqrt{65} }$ = - $\frac{-\sqrt{65} }{4}$

cosec θ = 1/sin θ = 1/ $\frac{\sqrt{65} }{7}$ = $\frac{-\sqrt{65} }{7}$

Learn more about trigonometric ratios: brainly.com/question/24349828

#SPJ1

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1 year ago