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BlackZzzverrR [31]
2 years ago
12

Ginas pet pot bellied pigs is on a diet. He can have no more than 18 ounces of pig food per day. How many scoops of pig food can

Gina feed the pig without going over 18 ounces?
PLS I NEED HELP ASAP
Mathematics
1 answer:
professor190 [17]2 years ago
6 0

Answer:

Step-by-step explanation:

Need to know how many ounces in a scoop of pig food. Is there a ratio or formula associated? Lmk.

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Mamont248 [21]

Here's some guidelines to help you:

For 1, 4, 5, and 6, they are 30-60-90 triangles. What's that you may ask? Well, it means that the sides have a predetermined length when one side is given. In #1 for example, you have a side of 11sqrt3, this means that "n" is 11 because in a 30-60-90 triangle, the longer side is the sqrt3 times the length of the shorter side. So to get the shorter side, we divide by sqrt3 to get 11. "m", or the hypotenuse, can be determined by taking twice the length of the shorter side. Since we figured out earlier that the shorter side is 11, 11 times 2 is 22. So the answer for #1 is n=11, m=22.

For 2 and 3, they are 45-45-90 triangles, triangles where the two lengths are the same and the hypotenuse is either leg length times sqrt2. In problem #2 for example, "y" must be 17 because one leg length is 17. "x", or the hypotenuse, is equal to 17sqrt(2) because 17 times sqrt(2).

You can apply all these rules to the other 4 problems I didn't explain.

Hope this long explanation clears your doubts!

5 0
3 years ago
The equation C = 5 9 F − 160 9 gives the relation between temperature readings in Celsius and Fahrenheit.
Pavel [41]

Step-by-step explanation:

The question is wrong. The correct equation is :

C=\frac{5}{9}F-\frac{160}{9}

We know that the equation gives the relation  between temperature readings in Celsius and Fahrenheit.

Therefore, giving that we know the value in Fahrenheit ''F'' we can find the reading in Celsius ''C''. This define a function C(F) that depends of the variable ''F''.

So for the incise (a) we answer Yes, C is a function of F.

For (b) we need to find the mathematical domain of this function. Giving that we haven't got any mathematical restriction, the mathematical domain of the function are all real numbers.

Dom (C) = ( - ∞ , + ∞)

For (c) we know that the water in liquid state and at normal atmospheric pressure exists between 0 and 100 Celsius.

Therefore the range will be

Rang (C) = (0,100)

Now, we need to find the domain for this range. We do this by equaliting and finding the value for the variable ''F'' :

For C = 0 :

0=\frac{5}{9}F-\frac{160}{9} ⇒ F=32

And for C = 100 :

100=\frac{5}{9}F-\frac{160}{9} ⇒ F=212

Therefore, the domain as relating temperatures of water in its liquid state is

Dom (C) = (32,212)

For (d) we only need to replace in the equation by F=71 and find the value of C ⇒

C=\frac{5}{9}F-\frac{160}{9} ⇒

C=(\frac{5}{9})(71)-\frac{160}{9}

C=\frac{65}{3} ≅ 21.67

C(71) = 21.67 °C

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