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3 years ago

Generating equivalent expressions

1 answer:

4 0

First you would solve what is in the parentheses so 3x-2x=1x or x. Then you would look to see if there are any like terms throughout the rest of the equation which there aren't. Therefore your answer is x+2y+2.

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1 — 3 (9x + 3) = 3x + 1

Hatshy [7]

Answer: x = -0.3

Step-by-step explanation:

Given expression

1 - 3 (9x + 3) = 3x + 1

Expand parentheses and apply the distributive property

1 - 27x - 9 = 3x + 1

Combine like terms

(1 - 9) - 27x = 3x + 1

-8 - 27x = 3x + 1

Subtract 3x on both sides

-8 - 27x - 3x = 3x + 1 - 3x

-8 - 30x = 1

Add 8 on both sides

-8 - 30x + 8 = 1 + 8

-30x = 9

Divide - 30 on both sides

(-30)x / (-30) = 9 / (-30)

(-30)x / (-30) = 3 / (-10)

$\boxed{x=-0.3}$

Hope this helps!! :)

Please let me know if you have any questions

3 0

3 years ago

Read 2 more answers

What is - 1 .2 % as a fraction and decimal

inysia [295]

-1 1/5 as a fraction -1.2 as a decimal

8 0

3 years ago

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Given the following diagram, find the missing measure. m<2=40°,m<3=150°,m<3=

Verdich [7]

Your answer be D and that is 30

5 0

3 years ago

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Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim xâ[infinity]

maksim [4K]

Answer:

$\frac{1}{5}$

Step-by-step explanation:

Given the expression;

$\lim_{x \to \infty} \dfrac{x+x^2}{5x^2}$

To find the limit of the function, we need to first divide through by the highest degree if x i.e x² as shown;

$\lim_{x \to \infty} \dfrac{\frac{x}{x^2} +\frac{x^2}{x2}}{\frac{5x^2}{x^2} }\\ \lim_{x \to \infty} \frac{\frac{1}{x}+ 1 }{5}\\As \ x \ tends \ to \ \infty, \ \frac{1}{x} \ tends \ to \ zero. \ Hence;\\ \lim_{x \to \infty} \frac{\frac{1}{x}+ 1 }{5} = \frac{0+1}{5}\\= \dfrac{1}{5}$

Hence the limit of the function as x tends to infinity is $\frac{1}{5}$

8 0

3 years ago

Simplify this expression (2c-7)(-7)

nevsk [136]

(2c-7)(-7)
= -24c+49

Hope this helps!

7 0

3 years ago