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defon
2 years ago
10

Question 12. Is the one I need help on

Mathematics
1 answer:
Mekhanik [1.2K]2 years ago
6 0

Four Answers:

  • Choice A
  • Choice C
  • Choice D
  • Choice J

====================================================

Explanation:

A rational number is any fraction of two whole numbers.

Examples: 7/9 and -8/12

This can be extended to some numbers like \sqrt{49} because

\sqrt{49} = \sqrt{7^2} = 7 = \frac{7}{1}

As you can see, any whole number is also rational but not the other way around.

So far, these examples help show that choices A, C, D, and J are rational numbers. Choice D might need a bit of rewriting like so:

\frac{0.5}{8} = 0.5 \div 8\\\\ = \frac{1}{2} \div \frac{1}{8} \\\\= \frac{1}{2} \times \frac{8}{1}\\\\ = \frac{8}{2}\\\\ = 4

In short, the expression for choice D simplifies to 4 = 4/1 showing that is rational. Or you can use the shortcut that rational/rational = rational.

------------------

A number like \sqrt{13} is NOT rational because the 13 is not a perfect square. There's no way to rewrite that square root as a ratio or fraction of integers. This allows us to rule out choice B. Choices E, F, and G are eliminated for similar reasoning.

Choice H is ruled out because circle areas involve pi = 3.14159... which is famously an irrational number. The proof of which is very lengthy to go over, but the short version is the decimal digits go on forever without a pattern. This is an informal way to check if we have an irrational number.

Choice I is only eliminated if and only if the side lengths of the square are rational, and also when the side lengths are whole numbers. Its not clear what your teacher's intentions here are, but I'll assume this is the case.

------------------

To summarize, we eliminated B, E, F, G, H, I

We're left with <u>A, C, D, J as the final answers</u>. They either can be written as fractions of integers, or already are fractions of integers.

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This question is down below. The picture attached below. <br> Thanks.
Ostrovityanka [42]

The vendor has to sell 88 gingerbread houses to earn a profit of $665.60 and there is no chance that the vendor will earn $1500.

Given an equation showing profits of A Christmas vendor as

P=-0.1g^{2}+30g-1200.

We have to find the number of gingerbread houses that the vendor needs to sell in order to earn profit of $665.60 and $1500.

To find the number of gingerbread houses we have to put P=665.60 in the equation given which shows the profit earned by vendor.

665.60=-0.1g^{2}+30g-1200

0.1g^{2}-30g+1200+665.60=0

0.1g^{2}-30g+1865.60=0

Divide the above equation by 0.1.

g^{2}-300g+18656=0

Solving for g we get,

g=[300±\sqrt{(300)^{2}-4*1*18656 }]/2*1

g=[300±\sqrt{90000-74624}]/2

g=[300±\sqrt{15376}]/2

g=(300±124)/2

g=(300+124)/2       , g=(300-124)/2

g=424/2,  g=176/2

g=212,88

Because 212 is much greater than 88 so vendor prefers to choose selling of 88 gingerbread houses.

Put the value of P=1500 in equation P=-0.1g^{2}+30g-1200.

-0.1g^{2}+30g-1200=1500

0.1g^{2}-30g+1500+1200=0

0.1g^{2}-30g+2700=0

Dividing equation by 0.1.

g^{2}-300g+27000=0

Solving the equation for finding value of g.

g=[300±\sqrt{300^{2} -4*1*27000}]/2*1

=[300±\sqrt{90000-108000}] /2

=[300±\sqrt{-18000}]/2

Because \sqrt{-18000} comes out with an imaginary number so it cannot be solved for the number of gingerbread houses.

Hence the vendor has to sell 88 gingerbread houses to earn a profit of $665.60 and there is no chance that the vendor will earn $1500.

Learn more about equation at brainly.com/question/2972832

#SPJ1

7 0
1 year ago
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