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2 years ago

Can anyone get this answer

Mathematics

1 answer:

Doss [256]2 years ago

7 0

This problem is asking you to apply the *Pythagorean Theorem*, given the information you’ve been given.

In case you’ve forgotten, the Pythagorean Theorem states that, in any given right triangle, the sum of the squares of the lengths of its legs is equal to the square of the length of its hypotenuse (the side opposite its right angle). If we call the lengths of the legs a and b, and the length of the hypotenuse c, this can be expressed in notation as a^2+b^2=c^2 (it doesn’t matter in this case which leg you pick for a and which you pick for b). Here, if we choose the left leg as a and the bottom leg as b, we’re given that a^2 (the area of a square with sides of length a) is 25 sq. in, and b is 3.5 in. Plugging those values into the equation, we have:

25 + (3.5)^2 = c^2

From here, you don’t even need to solve for c, you just need to find the value of c^2 (since you’re trying to find the area of a square with side lengths c). Just solve the left side of the equation, and you’ll have your answer in square inches.

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What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$