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3 years ago

Please correct answer

Mathematics

1 answer:

erica [24]3 years ago

3 0

Answer:

your right good job

Step-by-step explanation:

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Lin solved the equation 8(x-3)+7=2x(4-17) incorrectly. Find the errors in her solution. What should her answer have been?

Irina-Kira [14]

Answer: The correr solution is $x=\frac{17}{34}$

Step-by-step explanation:

Since the steps in Lin's solution are not shown, we are not able to find her erros. However, we can solve it step by step and find the correct result:

$8(x-3)+7=2x(4-17)$

Firstly, in the left part of the equation we have to multiply $8$ by the numbers inside the parenthesis; and in the right side we have to solve first what is inside the parenthesis:

$8x-24+7=2x(-13)$

Adding similar terms in the left side and multiplying $2x$ by the number inside the parenthesis in the right side:

$8x-17=-26 x$

Isolating $x$ :

$x=\frac{17}{34}$ This is the correct solution

4 0

3 years ago

For which of the following is x=-5 a solution. Select all that apply

mafiozo [28]

I think the answer is 3 but I’m not sure.

Hope this helps!

6 0

3 years ago

The ratio of girls to boys participating in intramural volleyball at Ashland Middle School is 7 to 4. There are 42 girls in the

LUCKY_DIMON [66]

There are 66 total participants

work:

6 0

3 years ago

Read 2 more answers

4x-2y=8(x-3) <br><br> x= ?<br> y= ?

Rainbow [258]

x = -6 y/2

y = -2x + 12

hope that helped

6 0

3 years ago

Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.

Shkiper50 [21]

Answer:

By closure property of multiplication and addition of integers,

If $x + \dfrac{1}{x}$ is an integer

∴ $\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

$x + \dfrac{1}{x}$

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}$

By simplification of the cube of the given integer expressions, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$

Therefore, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}$

By rearranging, we get;

$x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )$

Given that $x + \dfrac{1}{x}$ is an integer, from the closure property, the product of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3$ is an integer and $3\cdot \left (x + \dfrac{1}{x} \right )$ is also an integer

Similarly the sum of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

$\therefore x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer.

4 0

3 years ago