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Bad White [126]
3 years ago
13

3(5-8x) = 15 -4(6x - 1

Mathematics
1 answer:
vaieri [72.5K]3 years ago
6 0
   
\displaystyle \\
3(5-8x) = 15 -4(6x - 1) \\  \\ 
15-24x = 15 - 24x + 4 \\  \\ 
15-24x - 15 + 24x = 4 \\  \\ 
0 = 4 \\  \\ 
\Longrightarrow  ~~~\texttt{The equation has no solution.}



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In a graphing program, graph the following 4 functions in the same coordinate plane.
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See below for the changes when the exponential function is transformed

<h3>How to determine the effect of a</h3>

The exponential functions are given as:

y = \sqrt x

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y = 14\sqrt{x

y = -2\sqrt{x

An exponential function of the above form is represented as:

y = a\sqrt{x

See attachment for the graph of the four functions.

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This is represented by y = 14\sqrt{x

In this case, the curve of the base form y = \sqrt x is vertically stretched and it moves closer to the y-axis

<u>When a is small</u>

This is represented by y = 5\sqrt x

In this case, the curve of the base form y = \sqrt x is vertically stretched and it moves away from the x-axis

<u>When a is negative</u>

This is represented by y = -2\sqrt{x

In this case, the curve of the base form y = \sqrt x is vertically stretched and is reflected across the y-axis.

Read more about function transformation at:

brainly.com/question/26896273

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2 years ago
Graph the line. y=-4x+6 what the graph
Aloiza [94]
So the -4x is the M and then 6 is the B. So m is the slope and B is the y-intercept. The 6 goes on the y axis and the -4 will be how the line looks. It’ll be rise (-4) over run (1).

4 0
2 years ago
What values of a make the equation have non real solution?
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These can be some values of equation
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This question might not get the right answer but this how you set equations.






7 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E2%20%5Cleft%28%20%5Csqrt%5B3%5D%7B%20%7Bx%7D%5E%7B2%7D%20%20%2B%20
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Let f(x) = \sqrt{1 + x^3}. Compute the inverse:

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and we immediately notice that f^{-1}(x+1)=\sqrt[3]{x^2+2x}.

So, we can write the given integral as

\displaystyle \int_0^2 f^{-1}(x+1) + f(x) \, dx

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\displaystyle \int_1^3 f^{-1}(x) \, dx + \int_0^2 f(x) \, dx = 2 f(2) - 0 f(0) = 2\times3+0\times1=\boxed{6}

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Answer:

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Step-by-step explanation:

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