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3 years ago

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Someone please help me, this is dealing with the pythagorean theorem

1 answer:

Stella [2.4K]3 years ago

4 0

14. x ^2 = 12^2 + 7 ^2 = x = sqrt (12^2 + 7^2) = sqrt(193)

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Each Sunday, Mario weighs himself. Here are his weight changes over a 4-week period.

Dmitry [639]

So the answer is <span>–1.8 +0.9 –1.4 +0.4
or C. -1.9</span>

4 0

3 years ago

Read 2 more answers

If a square has a perimeter of 36 in. what is the Circumstances of the circle?given the exact answer in terms of pi.

vfiekz [6]

Answer:

a)B)9π

b)G)16π in²

Step-by-step explanation:

according to the question

$64 \times \frac{\pi}{4}$

$16\pi \: {in}^{2}$

as we know

C/d=π

C=dπ

Therefore

C=9π

6 0

3 years ago

IVE BEEN TRYING TO GET THE ANSWER FOR THIS WITH A PICTURE AND KEYBOARD PLEASE HELP (It’s a picture)

serg [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

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

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

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

3 years ago

In ΔIJK, i = 11 inches, j = 65 inches and k=70 inches. Find the area of AIJK to the nearest square inch.

earnstyle [38]

Answer:

330

Step-by-step explanation:

3 0

3 years ago