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allsm [11]
3 years ago
9

The quantity of 6 less than a number multiplied by 8

Mathematics
1 answer:
Bas_tet [7]3 years ago
5 0

6x-6=8x

subtract 6x from both sides

-6=2x

divide both sides by 2

x= -3

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From 1985 to 2007, the number B B of federally insured banks could be approximated by B ( t ) = − 329.4 t + 13747 B(t)=-329.4t+1
Romashka-Z-Leto [24]

Answer:

12100

Step-by-step explanation:

If the number B of federally insured banks could be approximated by B ( t ) = − 329.4 t + 13747 from 1985 to 2007 where t = 0 correspond to year 1985

In order to determine the amount of federally insured banks that were there in 1990, we will first calculate the year range from initial time 1985 till 1990

The amount of time during this period is 5years. Substituting t = 5 into the modeled equation will give;

B ( t ) = − 329.4 t + 13747

B(5) = -329.4(5) + 13747

B(5) = -1647+13747

B(5) = 12100

This shows that there will be 12100 federally insured banks are there in the year 1990.

5 0
3 years ago
Need help with this math​
Angelina_Jolie [31]

Answer:

205

Step-by-step explanation:

You divide the next answer which we presume is 55,555 by the constant 271 and get your final answer of 205.

Hope this helps bud:)

3 0
3 years ago
Read 2 more answers
931 is what percent of 980
Snowcat [4.5K]

Answer:

95%

Step-by-step explanation:

95% *980 =931

3 0
2 years ago
Elisa is making candles. She needs 5 ounces of wax for each candle. She has 25 ounces of wax. How many candles can she make?
Karolina [17]
25 divided by 5 = 5

The answer is 5 candles.
5 0
3 years ago
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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
2 years ago
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