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3 years ago

What is the solution to the equation log(2x + 4) = 2? Round to the nearest thousandth, if necessary.

Mathematics

1 answer:

Kazeer [188]3 years ago

5 0

Answer: 48

Step-by-step explanation:

$log_{b}M=n--> M=b^{n} \\log(2x+4)=2\\b=10, M=2x+4, n= 2 \\\\2x+4=10^{2} \\2x+4=100\\2x=96\\x=48$

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3 years ago

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3 years ago

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Darya [45]

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3 years ago

When the sum of \, 528 \, and three times a positive number is subtracted from the square of the number, the result is \, 120. F

aleksandr82 [10.1K]

Let $x$ be the unknown number. So, three times that number means $3x$ , and the square of the number is $x^2$

We have to sum 528 and three times the number, so we have $528+3x$

Then, we have to subtract this number from $x^2$ , so we have

$x^2-(3x+528)$

The result is 120, so the equation is

$x^2 - 3x - 528 = 120 \iff x^2 - 3x - 648 = 0$

This is a quadratic equation, i.e. an equation like $ax^2+bx+c=0$ . These equation can be solved - assuming they have a solution - with the following formula

$x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

If you plug the values from your equation, you have

$x_{1,2} = \dfrac{3\pm\sqrt{9-4\cdot(-648)}}{2} = \dfrac{3\pm\sqrt{9+2592}}{2} = \dfrac{3\pm\sqrt{2601}}{2} = \dfrac{3\pm51}{2}$

So, the two solutions would be

$x = \dfrac{3+51}{2} = \dfrac{54}{2} = 27$

$x = \dfrac{3-51}{2} = \dfrac{-48}{2} = -24$

But we know that x is positive, so we only accept the solution $x = 27$

6 0

3 years ago

In most microcomputers the addresses of memory locations are specified in hexadecimal. These addresses are sequential numbers th

iren [92.7K]

Considering that the addresses of memory locations are specified in hexadecimal.

a) The number of memory locations in a memory address range ( 0000₁₆ to FFFF₁₆ ) = 65536 memory locations

b) The range of hex addresses in a microcomputer with 4096 memory locations is ; 4095

<u>applying the given data </u>:

a) first step : convert FFFF₁₆ to decimal ( note F₁₆ = 15 decimal )

( F * 16^3 ) + ( F * 16^2 ) + ( F * 16^1 ) + ( F * 16^0 )

= ( 15 * 16^3 ) + ( 15 * 16^2 ) + ( 15 * 16^1 ) + ( 15 * 1 )

= 61440 + 3840 + 240 + 15 = 65535

∴ the memory locations from 0000₁₆ to FFFF₁₆ = from 0 to 65535 = 65536 locations

b) The range of hex addresses with a memory location of 4096

= 0000₁₆ to FFFF₁₆ = 0 to 4096

∴ the range = 4095

Hence we can conclude that the memory locations in ( a ) = 65536 while the range of hex addresses with a memory location of 4096 = 4095.

Learn more : brainly.com/question/18993173

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3 years ago