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Nata [24]
4 years ago
12

2k-6-7+3k = 48 need the answer!

Mathematics
2 answers:
GarryVolchara [31]4 years ago
5 0
The answer is k= 12 1/5
Radda [10]4 years ago
3 0

\huge{\fbox{\sf{ k= 12.2}}}

<u>Step by step explanation:</u>

Step 1:

2k-6-7+3k = 48

Step 2:

5k - 13 = 48

Step 3:

5k = 48 + 13

Step 4:

5k = 61

Step 6:

k = 12.2✓

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F(n) = n² – 3 g(n) = 4n - 1 Find f[g(1)]
SVETLANKA909090 [29]

Answer:

f[g(1)]=6.

Explanation:

Given f(n) and g(n) defined below:

\begin{gathered} f\mleft(n\mright)=n^2-3 \\ g\mleft(n\mright)=4n-1 \end{gathered}

First, we evaluate g(1):

\begin{gathered} g\mleft(1\mright)=4(1)-1 \\ =4-1 \\ g(1)=3 \end{gathered}

Therefore:

\begin{gathered} f\mleft(g(1)\mright)=f\mleft(3\mright) \\ f\mleft(3\mright)=3^2-3 \\ =9-3 \\ =6 \end{gathered}

Therefore, f[g(1)]=6.

8 0
1 year ago
In a certain game of chance, your chances of winning are 0.3. Assume outcomes are independent and that you will play the game fo
sergeinik [125]

Answer:

0.6517

Step-by-step explanation:

Given that in a certain game of chance, your chances of winning are 0.3.

We know that each game is independent of the other and hence probability of winning any game = 0.3 (constant)

Also there are only two outcomes

Let X be the number of games you win when you play 4 times

Then X is binomial with p = 0.3 and n =4

Required probability

= Probability that you win at most once

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We have as per binomial theorem

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3 0
3 years ago
Match the expressions with their equivalent simplified expressions.
Tasya [4]

Answer:

\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}


Step-by-step explanation:

\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}} =\sqrt[4]{\frac{(2^4)(x^{6-2})(y^{4-8})}{(3^4)}} =\sqrt[4]{\frac{2^4x^4y^{-4}}{3^4}} =\frac{2xy^{-1}}{3}=\frac{2x}{3y}

\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} =\sqrt[4]{\frac{(3^4)(x^{2-6})(y^{10-6})}{(2^4)}} =\sqrt[4]{\frac{3^4x^{-4}y^{4}}{2^4}} =\frac{3x^{-1}y^1}{3}=\frac{3y}{2x}

\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}} =\sqrt[3]{\frac{(4^3)(x^{8-2})(y^{7-10})}{(5^3)}} =\sqrt[3]{\frac{4^3x^6y^{-3}}{5^3}} =\frac{4x^2y^{-1}}{5}=\frac{4x^2}{5y}

\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}} =\sqrt[5]{\frac{(3^5)(x^{17-7})(y^{16-21})}{(2^5)}} =\sqrt[5]{\frac{3^5x^{10}y^{-5}}{2^5}} =\frac{3x^2y^{-1}}{2}=\frac{3x^2}{2y}

\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} =\sqrt[5]{\frac{(2^5)(x^{12-7})(y^{15-10})}{(3^5)}} =\sqrt[5]{\frac{2^5x^{5}y^{5}}{3^5}} =\frac{2x^1y^{1}}{3}=\frac{2xy}{3}

\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}} =\sqrt[4]{\frac{(2^4)(x^{10-2})(y^{9-17})}{(4^4)}} =\sqrt[4]{\frac{2^4x^{8}y^{-8}}{4^4}} =\frac{2x^{1}y^{-1}}{4}=\frac{x}{2y}

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\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}

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