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3 years ago

What is the approximate solution of e2x−3 = 7?

Mathematics

2 answers:

evablogger [386]3 years ago

8 0

E2x-3=7 (multiply by ln e on both sides)
ln e of e=1
>>> 2x-3=ln 7
x=(ln 7 + 3)/2
x=2.47 (to 3 s.f)

Harlamova29_29 [7]3 years ago

3 0

The approximate solution is .69314718

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There is a bag filled with 4 blue, 3 red and 5 green marbles. A marble is taken at random from the bag, the colour is noted and

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The exponential function, f(x)=2^x, undergoes two transformations to g(x)=3x2^x+5. How does the graph change? Select all that ap

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Answer: Options A and C.

Step-by-step explanation:

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3 years ago

Six boys are comparing their heights. Andy is taller than Dan but shorter than Ethan. Bill is taller than Ethan and Frank. Frank

Alexxx [7]

List of boys from tallest to shortest is: Bill, Frank, Ethan, Andy, Dan, Cory

<h3>Solution:</h3>

Given that Six boys are comparing their heights

The six boys are Andy, Dan, Ethan, Bill, Frank, Cory

To find: List the boys from tallest to shortest

Andy is taller than Dan but shorter than Ethan

So the arrangement goes like this:

Ethan

Andy

Dan

Bill is taller than Ethan and Frank:

So Bill is taller than ethan. Thus the arrangement from tall to short is made as:

Bill

Ethan

Andy

Dan

Also bill is taller than frank. So frank can be placed anywhere below bill

But given that Frank is taller than Ethan. So frank can be placed above ethan

Now the arrangement goes like this:

Bill

Frank

Ethan

Andy

Dan

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So cory can be placed below Dan.

Thus final arrangement from tallest to shortest boys are:

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3 years ago

Find the product of all real values of r for which 1/2x=r-x/7

Dahasolnce [82]

Answer:

$r = \±\sqrt{14$

$Product = -14$

Step-by-step explanation:

Given

$\frac{1}{2x} = \frac{r - x}{7}$

Required

Find all product of real values that satisfy the equation

$\frac{1}{2x} = \frac{r - x}{7}$

Cross multiply:

$2x(r - x) = 7 * 1$

$2xr - 2x^2 = 7$

Subtract 7 from both sides

$2xr - 2x^2 -7= 7 -7$

$2xr - 2x^2 -7= 0$

Reorder

$- 2x^2+ 2xr -7= 0$

Multiply through by -1

$2x^2 - 2xr +7= 0$

The above represents a quadratic equation and as such could take either of the following conditions.

(1) No real roots:

This possibility does not apply in this case as such, would not be considered.

(2) One real root

This is true if

$b^2 - 4ac = 0$

For a quadratic equation

$ax^2 + bx + c = 0$

By comparison with $2x^2 - 2xr +7= 0$

$a = 2$

$b = -2r$

$c =7$

Substitute these values in $b^2 - 4ac = 0$

$(-2r)^2 - 4 * 2 * 7 = 0$

$4r^2 - 56 = 0$

Add 56 to both sides

$4r^2 - 56 + 56= 0 + 56$

$4r^2 = 56$

Divide through by 4

$r^2 = 14$

Take square roots

$\sqrt{r^2} = \±\sqrt{14$

$r = \±\sqrt{14$

Hence, the possible values of r are:

$\sqrt{14$ or $-\sqrt{14$

and the product is:

$Product = \sqrt{14} * -\sqrt{14}$

$Product = -14$

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3 years ago