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erastova [34]
3 years ago
13

I notice my dog has gained some weight and so i decided to reduce the amount of food that I feed him daily. I decide to start fe

eding him 1 1/10 pounds of food every day instead of the 1 7/8 pounds that I was feeding him daily. how much less dog food am I going to need in a month? (write your answer as a mixed number)
Mathematics
1 answer:
nirvana33 [79]3 years ago
5 0
Well, first you would want to turn 1 1/10 and 1 7/8 into improper fractions. This would result in 11/10 and 15/8.

Next you want to find the smallest common denominator:
11/10 (4/4) = 44/40
15/8 (5/5) = 75/40

You would then subtract:
75/40 - 44/40 = 31/40

The number 31/40 would be the amount of food you aren't giving the dog per day (are cutting back on). Now you jut have to find out how many days are in the month to get your final answer.
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Which expression is a fourth root of -1+isqrt3?
aleksklad [387]

Answer:

Step-by-step explanation:

\sf n^{th} roots of a complex number is given by DeMoivre's formula.

   \sf \boxed{\bf r^{\frac{1}{n}}\left[Cos \dfrac{\theta + 2\pi k}{n}+i \ Sin \ \dfrac{\theta+2\pi k}{n}\right]}

Here, k lies between 0 and (n -1) ; n is the exponent.

\sf -1 + i\sqrt{3}

a = -1 and b = √3

\sf \boxed{r=\sqrt{a^2+b^2}} \ and \ \boxed{\theta = Tan^{-1} \ \dfrac{b}{a}}

\sf r = \sqrt{(-1)^2 + 3^2}\\\\ = \sqrt{1+9}\\\\=\sqrt{10}

                   \sf \theta = tan^{-1} \ \dfrac{\sqrt{3}}{-1}\\\\ = tan^{-1} \ (-\sqrt{3})

                   \sf = \dfrac{-\pi }{3}

n = 4

For k = 0,

          \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin  \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi  }{12}+iSin  \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]

For k =1,

         \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{5\pi}{12}+i \ Sin \ \dfrac{5\pi}{12}\right]

For k =2,

       z = \sqrt[4]{10}\left[Cos \ \dfrac{11\pi}{12}+i \ Sin \ \dfrac{11\pi}{12}\right]

For k = 3,

      \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{17\pi}{12}+i \ Sin \ \dfrac{17\pi}{12}\right]

For k = 4,

      \sf z =\sqrt[4]{10}\left[Cos \ \dfrac{23\pi}{12}+i \ Sin \ \dfrac{23\pi}{12}\right]

4 0
1 year ago
What is the solution of 9|2x – 1| + 4 < 49?
luda_lava [24]

Answer: -2

Step-by-step explanation:

You need to set up two cases (Positive case and negative case) and solve for "x".

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9(2x -1) + 4 < 49\\18x-9

- NEGATIVE CASE IF: 2x-1

-9(2x -1) + 4 < 49\\-18x+9

Therefore, the solution is:

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