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3 years ago

ABC is a straight line. which equation can be used to find the measure of the unknown angel?

Mathematics

1 answer:

miskamm [114]3 years ago

7 0

Answer:

Step-by-step explanation:

Since it is a straight angle, it will equal 180 degrees. Since 45 is already taken, u would subtract it from 180.

180-45=135

thus, your answer is 180-45

hope I helped!!!

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Solve for an angle in right triangles

Aneli [31]

Answer:

Step-by-step explanation:

if all the interior angle in triangle sums up it will be 180

so 180 divide by by 3(because its a triangle) is 90

hope this helps

have a nice day <3

8 0

2 years ago

I need help so I can pass this plz ASAP

4vir4ik [10]

Answer:

-5

Step-by-step explanation:

3 0

3 years ago

I need help please?!!!

Sveta_85 [38]

Answer:

(-4,14)

(which you already choose)

Step-by-step explanation:

5 0

3 years ago

What’s the equation for the order pairs (1,-5) and (-2,5)

liq [111]

$\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{-5})\qquad
(\stackrel{x_2}{-2}~,~\stackrel{y_2}{5})
\\\\\\
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{5-(-5)}{-2-1}\implies \cfrac{5+5}{-2-1}\implies \cfrac{10}{-3}$

$\bf \begin{array}{|c|ll}
\cline{1-1}
\textit{point-slope form}\\
\cline{1-1}
\\
y-y_1=m(x-x_1)
\\\\
\cline{1-1}
\end{array}\implies y-(-5)=-\cfrac{10}{3}(x-1)\implies y+5=-\cfrac{10}{3}x+\cfrac{10}{3}
\\\\\\
y=-\cfrac{10}{3}x+\cfrac{10}{3}-5\implies y=-\cfrac{10}{3}x-\cfrac{5}{3}$

6 0

3 years ago

Expand (2x+2)^6<br> How would you find the answer using the binomial theorem?

Yanka [14]

Answer:

Step-by-step explanation:

$\displaystyle\\\sum\limits _{k=0}^n\frac{n!}{k!*(n-k)!}a^{n-k}b^k .\\\\k=0\\\frac{n!}{0!*(n-0)!}a^{n-0}b^0=C_n^0a^n*1=C_n^0a^n.\\\\ k=1\\\frac{n!}{1!*(n-1)!} a^{n-1}b^1=C_n^1a^{n-1}b^1.\\\\k=2\\\frac{n!}{2!*(n-2)!} a^{n-2}b^2=C_n^2a^{n-2}b^2.\\\\k=n\\\frac{n!}{n!*(n-n)!} a^{n-n}b^n=C_n^na^0b^n=C_n^nb^n.\\\\C_n^0a^n+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+...+C_n^nb^n=(a+b)^n.$

$\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\$

$(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+ \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.$

8 0

1 year ago