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2 years ago

Margie received her store order on 12/3/16 at 4AM. She just opened one of the

Mathematics

1 answer:

Cerrena [4.2K]2 years ago

8 0

Answer:

Exp Date: 1/17/2017

Exp Time: 4:00am

Prep Date: 12/3/2016

Prep Time: 4:00am

Initials: 12/7/2016

Step-by-step explanation:

A well-detailed version of the question has been uploaded in form of an image for easier understand.

Looking at the question, it was said that she received her store order on 12/3/2016 at 4am, this implies that the prep date and time are 12/3/2016 and 4am respectively. Also, it was said that the expiration date was printed on the product and it is 1/17/2017. Obviously, the expiration time would also be 4am because it was prepared at 4am and if we calculate in. 24hours we would get 4am at the expiration date as well. Lastly, we were told she opened the product she received on 12/7/2016 which is the initial date the product was used. From all these we can deduce the following:

Exp Date: 1/17/2017

Exp Time: 4:00am

Prep Date: 12/3/2016

Prep Time: 4:00am

Initials: 12/7/2016

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Find a right triangle that has legs with your irrational Lengths and a hypotenuse with a rational Length

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gg easy

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let's find some stuff

one easy way is to find a rational value for c find values of a and b

lets pick c=5

$a^2+b^2=5^2$

$a^2+b^2=25$

now split up 25 into 2 things that don't have rational square roots

if we say $a^2=19$ and $b^2=6$ then we get

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2 years ago

Please answer all of them need this

VikaD [51]

First Question

For a better understanding of the solution provided here please find the first attached file which has the diagram of the the isosceles trapezoid.

We dropped perpendiculars from C and D to intersect AB at Q and P respectively.

As can be seen in $\Delta BCQ$ , we can easily find the values of CQ and BQ.

Since, $Sin(75^0)=\frac{CQ}{8}$

$\therefore CQ=8\times Sin(75^0)\approx 7.73$ ft

In a similar manner we can find BQ as:

$Cos(75^0)=\frac{BQ}{8}$

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All these values can be found in the diagram attached.

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Let us now consider $\Delta AQC$

We can apply the Pythagorean Theorem here to find the length of the diagonal AC which is the hypotenuse of $\Delta AQC$ .

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Thus, out of the given options, Option B is the closest and hence is the answer.

Second Question

For this question we can directly apply the formula for the area of a triangle using sines which is as:

Area= $\frac{1}{2}$ (First Side)(Second Side)(Sine of the angle between the two sides)

Thus, from the given data,

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Therefore, Option D is the correct option.

Third Question

For this question we will apply the Sine Rule to the $\Delta ABC$ given to us.

Thus, from the triangle we will have:

$\frac{AB}{Sin(\angle C)}=\frac{BC}{Sin(\angle A)}$

$\frac{c}{Sin(\angle C)}=\frac{a}{Sin(\angle A)}$

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This gives $a$ to be:

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Which is not close to any of the given options.

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Please find the second attachment for a better understanding of the solution provided her.

As can be clearly seen from the attached diagram, we can apply the Cosine Rule here to find the return distance of the plane which is CA.

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