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nekit [7.7K]
4 years ago
7

PLEASE HELP I AM STUCK ON THIS

Mathematics
1 answer:
Fofino [41]4 years ago
4 0

Answer:

Step-by-step explanation:

This is a right triangle problem.  The reference angle is x, the side opposite the reference angle is 32, and the hypotenuse is 58.  The trig ratio that relates the side opposite a reference angle to the hypotenuse is the sin.  Filling in accordingly:

sin(x)=\frac{32}{58}

Because you are looking for a missing angle, you will use your 2nd button and then the sin button to see on your display:

sin^{-1}(

Within the parenthesis enter the 32/58 and you'll get your angle measure.  Make sure your calculator is in degree mode, not radian mode!!!

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6.68754...

Step-by-step explanation:

Assuming you mean:

\sqrt{2}\cdot 3^{\sqrt{2}}

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25t - 87.5 = 12.5<br> What is the answer
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Step-by-step explanation:

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What is 37/50 as a decimal
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(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
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