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3 years ago

Find the missing side lengths leave your answerss as radicals??

Mathematics

1 answer:

ASHA 777 [7]3 years ago

4 0

Answer:

Step-by-step explanation:

1) Add 30+90 to get 120 and subtract 180-120= 60 as your third angle.

2) Take sin60= x/6 to get x= 3√3

3) Take sin30= y/6 to get y= 3

Here's a tip to verify this answer.

Use Pythagorean Theorem to do:

Square root of (3√3)^2 + 3^2 which equals 6 as your hypotenuse!

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Whats the answer to this i need help please

Elan Coil [88]

51 feet you have to multiple 32 by 8 it will give you 256 and then divide 256 and 5 and you get 51

3 0

3 years ago

Given f(x) = x − 7 and g(x) = x2 . Find g(f(−1)).

ivann1987 [24]

Answer:

Step-by-step explanation:

$g(f(-1))$ can be found by first finding $f(-1)$ then taking that result and putting it into the function $g$ .

$f(-1)$ means we are going to take the expression $x-7$ and evaluate it for $x=-1$ :

$-1-7$

$-8$

So $f(-1)=-8$ .

So we have this so far:

$g(f(-1))=g(-8)=(-8)^2=64$ .

$g(-8)$ was found by replacing $x$ in $x^2$ with $-8$ .

8 0

3 years ago

Find y when x=12 if y=15 when x=3

larisa [96]

The answer would be y=60.

I think.

4 0

3 years ago

Josephine makes $29,695.06 annually, and she spends $5,048.16 per year on Food. What percent of her income goes towards Food?

Igoryamba

Answer:

17%

Step-by-step explanation:

$5,048.16 divided by $29,695.06 = 0.16999999

Move the decimal over 2 places to the right to get a percentage.

16.99999 = 17%

7 0

3 years ago

Using the pattern, give the coefficients of (x + y)^5 and (x + y)^6

musickatia [10]

Answer:

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

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$

Step-by-step explanation:

In order to find the values of $(x+y)^5$ and $(x+y)^6$ , you need to apply the binomial theorem (high-level math you most likely don't need to worry about, it's easier than multiplying all the binomials together).

$(x+y)^5$ = $\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$ = $\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$ = $x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$ .

$(x+y)^6$ = $\sum _{i=0}^6\binom{6}{i}x^{\left(6-i\right)}y^i$ = $\frac{6!}{0!\left(6-0\right)!}x^6y^0+\frac{6!}{1!\left(6-1\right)!}x^5y^1+\frac{6!}{2!\left(6-2\right)!}x^4y^2+\frac{6!}{3!\left(6-3\right)!}x^3y^3+\frac{6!}{4!\left(6-4\right)!}x^2y^4+\frac{6!}{5!\left(6-5\right)!}x^1y^5+\frac{6!}{6!\left(6-6\right)!}x^0y^6$ = $x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$ .

5 0

3 years ago