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2 years ago

3(2x-7=5-(1-x) pls help me

Mathematics

1 answer:

vekshin12 years ago

7 0

Answer:

3 • (2x - 7) - (x + 4) = 0

then do 3.1 Pull out like factors :

5x - 25 = 5 • (x - 5)

5 • (x - 5) = 0

4.2 Solve : x-5 = 0

Add 5 to both sides of the equation :

x = 5

Step-by-step explanation:

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A sample pf a radioactive isotope had an initial mass of 370 mg in the year 2007 and

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Answer:

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Step-by-step explanation:

8 0

1 year ago

Which pair of numbers are not opposites?

mrs_skeptik [129]

Answer:

The answer is C, 17 and | -17|

Step-by-step explanation:

We can solve this using the process of elimination.

Looking at the first one, we know that those | | around the 7 mean to find the absolute value. The absolute value is the number of units the number is from zero, and it is always positive. This means that it is saying 7 and -7. Those are opposites, so we know that this isn't right.

17 and -17 are opposites, since the opposite of a positive is a negative.

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2 years ago

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Pre cal/cal master needed

weqwewe [10]

Take x-2 and insert it into 2x^2 + 3x-2 where the x is located
2x^2 + 3x-2
2(x-2)^2 + 3(x-2)-2

Now work out 2(x-2)^2 + 3(x-2)-2 also follow PEMDAS
2(x-2)^2 + 3(x-2)-2
Since (x-2)^2 is an Exponent, lets work with that first and expand (x-2)^2.
(x-2)^2
(x -2)(x-2)
x^2 -4x + 4
Now Multiply that by 2 because we have that in 2(x-2)^2
(x-2)^2 = x^2 -4x + 4
2(x-2)^2 = 2(x^2 -4x + 4)
2(x^2 -4x + 4) = 2x^2 - 8x + 8
2x^2 - 8x + 8

Now that 2(x-2)^2 is done lets move on to 3(x-2).

Use the distributive property and distribute the 3
3(x-2) = 3x - 6

All that is left is the -2

Now lets put it all together
2(x-2)^2 + 3(x-2)-2

2x^2 - 8x + 8 + 3x - 6 - 2

Now combine all our like terms
2x^2 - 8x + 8 + 3x - 6 - 2
Combine: 2x^2 = 2x^2
Combine: -8x + 3x = -5x
Combine: 8 - 6 - 2 = 0

So all we have left is
2x^2 - 5x

5 0

3 years ago

Find the domain of the function y = 3 tan(23x)

solmaris [256]

Answer:

$\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

In other words, the $x$ in $f(x) = 3\, \tan(23\, x)$ could be any real number as long as $x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)$ for all integer $k$ (including negative integers.)

Step-by-step explanation:

The tangent function $y = \tan(x)$ has a real value for real inputs $x$ as long as the input $x \ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

Hence, the domain of the original tangent function is $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

On the other hand, in the function $f(x) = 3\, \tan(23\, x)$ , the input to the tangent function is replaced with $(23\, x)$ .

The transformed tangent function $\tan(23\, x)$ would have a real value as long as its input $(23\, x)$ ensures that $23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

In other words, $\tan(23\, x)$ would have a real value as long as $x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right)$ .

Accordingly, the domain of $f(x) = 3\, \tan(23\, x)$ would be $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

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2 years ago

The approximate populations of Massachusetts and Wyoming (as of 2019) are listed

chubhunter [2.5K]

First can you tell me what the answer to the 6.89 * 106 for Massachusetts is

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2 years ago

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