The first occurrence of the highest temperature in the

Which answer is correct? It wouldn’t let me click mathematics btw

Bad White [126]

6)
KM bisects on <NKL so m<MKN = m<MKL = 3x + 5

2(m<MKL) + m<JKN = 180
2(3x+5) + 8x + 2 = 180
6x + 10 + 8x + 2 = 180
14x + 12 = 180
14x = 168
x = 12

so
m<MKN = m<MKL = 3x + 5 = 3(12) + 5 = 36 + 5 = 41

answer
b. 41

7)

2 supplementary has sum = 180
x + x + 14 = 180
2x + 14 = 180
2x = 166
x = 83
x + 14 = 83 + 14 = 97

2 angles are 83 and 97

answer
b. 83, 97

7 0

4 years ago

The answers to these questions please

vfiekz [6]

Answer:

Q10 is 13.64 cm3

Q 11 is 221 cm3

Step-by-step explanation:

volume is cross sectional area times height for Q 10

volume is length times width times height for Q 11

5 0

3 years ago

PLEASE HELP!

Aleksandr [31]

Answer:8x10^-8

Step-by-step explanation:I am brainlyest

3 0

3 years ago

A graph of a system of two equations is shown. HELP!!!!!

s344n2d4d5 [400]

Answer:

8 and 2

Step-by-step explanation:

8 and 2 because y=5x +3 would be 5+3 which is 8 and y= 3x - 1 would be 3 - 1 which equals 2.

4 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

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

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

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

$=5x^4y$

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