20 POINTS

Name: 11. Find the measure of each acute angle. (2x + 6° (5x + 14°

Westkost [7]

Answer:

A

Step-by-step explanation:

Dc is a little bit more than a few other things but I think it’s Going and the kids have to be in a different room yet I think it’s going on for me and my kids are in my car so I’m going to go get my car off at my moms and I can go pick it out I have a lot to go

4 0

3 years ago

Read 2 more answers

Subtract fractions like 42/7- 2 1/5

dangina [55]

Answer:

3.8 in decimal form, or 3 4/5

Step-by-step explanation:

I just used a calculator, sorry :(

anyways, hope this helped! :)

6 0

3 years ago

kenny6666 [7]

△PQR is similar to △STU

m∠R = m∠U = 96°

m∠Q = m∠T = 6

m∠P = 180 - ( 96 + 6)

m∠P = 180 - 102

m∠P = 78

answer

m∠P = 78

7 0

3 years ago

Read 2 more answers

Maris put trim around a mirror that is the shape of a square. Each side is 24 inches long. Maria has 1 foot of trim left. What w

Dahasolnce [82]

Answer:

16.028 yards

Step-by-step explanation:

1 inch=0.0833333 foot

24 inches=0.0833333×24

=1.9999992 foot

The mirror is square shaped and each side is 24 inches

If each 24 inches=1.9999992 foot

Then,

4 sides of 24 inches=1.9999992×24

=47.9999808 foot

Maria has 1 foot of trim left

Total Length of trim when she started=47.9999808+0.0833333

=48.0833141 foot

=16.02777136667 yards

Approximately,

16.028 yards

8 0

3 years ago

(x +y)^5 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

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

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

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

$=5x^4y$

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

2 years ago