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3 years ago

22 more than four times a number is less than 82 (Inequality equation)

Mathematics

1 answer:

Tanzania [10]3 years ago

5 0

22^4c-82. Is the equation I think you are looking for.

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What is the answer to this equation |x−1|+5=2

barxatty [35]

Answer:

No solutions

Step-by-step explanation:

Isolate the absolute value:

|x−1| + 5 = 2

Subtract 5 from both sides:

|x-1| = -3

Since an absolute value can never be equal to a negative number, there are no solutions.

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2 years ago

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3 years ago

What is the fourth term in the binomial expansion (a+b)^6)

Dafna11 [192]

Answer:

$20a^3b^3$

Step-by-step explanation:

<u>Binomial Series</u>

$(a+b)^n=a^n+\dfrac{n!}{1!(n-1)!}a^{n-1}b+\dfrac{n!}{2!(n-2)!}a^{n-2}b^2+...+\dfrac{n!}{r!(n-r)!}a^{n-r}b^r+...+b^n$

<u>Factorial</u> is denoted by an exclamation mark "!" placed after the number. It means to multiply all whole numbers from the given number down to 1.

Example: 4! = 4 × 3 × 2 × 1

Therefore, the fourth term in the binomial expansion (a + b)⁶ is:

$\implies \dfrac{n!}{3!(n-3)!}a^{n-3}b^3$

$\implies \dfrac{6!}{3!(6-3)!}a^{6-3}b^3$

$\implies \dfrac{6!}{3!3!}a^{3}b^3$

$\implies \left(\dfrac{6 \times 5 \times 4 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}{3 \times 2 \times 1 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}\right)a^{3}b^3$

$\implies \left(\dfrac{120}{6}\right)a^{3}b^3$

$\implies 20a^3b^3$

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2 years ago