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3 years ago

Choose the correct simplification of the expression (3x − 6)(2x2 − 4x − 5).

Mathematics

2 answers:

zavuch27 [327]3 years ago

8 0

The answer is C: 6x^3 - 24x^2 + 9x + 30

Deffense [45]3 years ago

8 0

Answer:

Option 3rd is correct

$6x^3-24x^2+9x+30$

Step-by-step explanation:

Given the expression:

$(3x-6)(2x^2-4x-5)$

Multiply the first term of the first expression with second expression and second term of the first expression with second expression we have;

$3x(2x^2-4x-5)-6(2x^2-4x-5)$

using distributive property we have;

$6x^3-12x^2-15x-12x^2+24x+30$

Combine like terms we have;

$6x^3-24x^2+9x+30$

therefore, the correct simplification of the given expression is, $6x^3-24x^2+9x+30$

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A local gym had 490 members last quarter. This quarter, with a change in its rates, it has 686 members. What is the percent of i

sergij07 [2.7K]

Given:

Number of gym members in last quarter = 490

Number of gym members in this quarter = 686

To find:

The percent of increase in the number of members.

Solution:

We have,

Number of gym members in last quarter = 490

Number of gym members in this quarter = 686

Now, the percent of increase in the number of members.

$\%\text{ of increase}=\dfrac{{Members in this month - Members in last month}}{\text{Members in last month}}\times 100$

$\%\text{ of increase}=\dfrac{686-490}{490}\times 100$

$\%\text{ of increase}=\dfrac{196}{490}\times 100$

$\%\text{ of increase}=\dfrac{2}{5}\times 100$

$\%\text{ of increase}=2\times 20$

$\%\text{ of increase}=40$

Therefore, the gym members increased by 40%.

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2 years ago

Which relation is not a function? Reporting wrong answers.

Aleks [24]

Answer:

C im pretty sure dont report me tho

Step-by-step explanation:

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2 years ago

Determine whether the series is convergent or divergent by expressing sn as a telescoping sum (as in Example 8). [infinity] 6 n2

vekshin1

Answer:

Step-by-step explanation:

given

$\sum\limits^\infty_{n=3}\frac{6}{n^2-1}=\sum\limits^\infty_{n=3}\frac{6}{(n-1)(n+1)}\\\\=\sum\limits^\infty_{n=3}\frac{6}{2}(\frac{1}{n-1}-\frac{1}{n+1})\\\\=\frac{6}{2}\sum\limits^\infty_{n=3}(\frac{1}{n-1}-\frac{1}{n+1})\\\\=3[(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+(\frac{1}{5}-\frac{1}{7})+...]\\\\=3[\frac{1}{2}+\frac{1}{3}]=3[\frac{3+2}{6}]=3[\frac{5}{6}]\\\\=\frac{15}{6}=\frac{5}{3}$