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4 years ago

What is the value of x in the equation (x + 12) = (x + 14) – 3?

2 answers:

6 0

(x + 12) = (x + 14) - 3

subtract 12 from each side

x = x + 14 - 3 - 12

Now if we subtract x from each side we see that x - x = 0. Combine the terms left over and we have -1, so

0 = -1

The answer is NO SOLUTION

deff fn [24]4 years ago

6 0

Answer: There is NO value of 'x' in the given equation.

Step-by-step explanation: We are given to find the value of 'x' in the equation below:

$(x+12)=(x+14)-3~~~~~~~~~~~~~~~~~~~~(i)$

Finding the value of 'x' in equation (i) means to find the value of 'x' for which the equation is TRUE.

The solution is as follows:

$(x+12)=(x+14)-3\\\\\Rightarrow x+12=x+14-3\\\\\Rightarrow x+12=x+11\\\\\Rightarrow x-x=11-12\\\\\Rightarrow 0=-1,$

which can NEVER be true.

So, the given equation has NO solution.

Thus, there is NO value of 'x' in the given equation.

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(07.03. 07.04 MC)

posledela

Answer:

A) 4x^2+20x+25=(2x)^2+2*(2x)*5+5^2=(2x+5)^2

Area=(side)^2, side=sqrt(area)=sqrt((2x+5)^2)=2x+5

B) 4x^2-9y^2=(2x-3y)(2x+3y), these are the dimensions of the rectangle

8 0

3 years ago

hey certain cars value is said to depreciate by approximately 15% each year for the first eight years of the car originally cost

soldi70 [24.7K]

Answer:0.0625

Step-by-step explanation:

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4 years ago

HELPPP MEEE PLEASEE!!!!!

Sergio [31]

For this case we have that, by definition, the domain of a function is given by all the values for which the function is defined.

Having said that, we have the following function:

$r (x) = \frac {2x} {x-1}$

It is not defined when the denominator is equal to 0. That is:

$x-1 = 0\\x = 0 + 1\\x = 1$

Thus, the domain of r (x) is given by all real numbers except 1.

Answer:

all real numbers except 1.

Option A

6 0

4 years ago

X = 10 x = 2 x = two divided by five. x = 11

KengaRu [80]

Answer:

x = 10

Step-by-step explanation:

You can try the answers to see which works. (The first one does.)

Or, you can solve for the variable:

Divide by 75

... (1/5)^(x/5) = 3/75 = 1/25

Recognize that 25 = 5^2, so ...

... (1/5)^(x/5) = (1/5)^2

Equating exponents, you have

... x/5 = 2

... x = 10 . . . . . multiply by 5

_____

You can also start by taking logarithms:

... log(75) +(x/5)log(1/5) = log(3)

... (x/5)log(1/5) = log(3) -log(75) = log(3/75) = log(1/25) . . . . simplify the log

... x/5 = log(1/25)/log(1/5) = 2 . . . . . simplify (or evaluate) the log expression

... x = 10 . . . . . multiply by 5

_____

"Equating exponents" is essentially the same as taking logarithms.

5 0

3 years ago

Please help evaluate?

Dominik [7]

3.

4^3 is equal to 64. Therefor log4(64) must be equal to 3.

3 0

3 years ago