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3 years ago

9

DG¯¯¯¯¯¯ , EG¯¯¯¯¯ , and FG¯¯¯¯¯ are perpendicular bisectors of the sides of △ABC . DG=5 centimeters and BD=12 centimeters. What

is CG ? Enter your answer in the box.

2 answers:

pshichka [43]3 years ago

4 0

Your answer should be 13cm

AysviL [449]3 years ago

4 0

The answer should be 13 cm, I'm taking the same quiz.

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Please Help Me With This.

borishaifa [10]

$\rule{50}{1}\large\textsf{\textbf{\underline{Question:-}}}\rule{50}{1}$

<em>[refer to attachment for question] ~</em>

<em> </em>

<em> </em> $\rule{50}{1}\large\blue\textsf{\textbf{\underline{Answer and how to solve:-}}}\rule{50}{1}$

✫ First of all, Slope-intercept form is written as follows:-

$\longmapsto\Large\textbf{y=mx+b}$

<u>Where:-</u>

m = slope ;
b = y-intercept

<u>To find:-</u>

<u></u>

The equation of the line

<u>Solving,</u>

Replace m with -1/5 And b with -3:-

$\large\text{$y=-\displaystyle\frac{1}{5} x-3$}$

$\Uparrow$ $\sf{the\:equation}$

<h3>Good luck with your studies. $\ddot\smile$ </h3>

#TogetherWeGoFar

$\rule{300}{1}$

4 0

2 years ago

6^x = 216 how to solve? 6 to the power of x equals 216.

Nataly_w [17]

$6 \cdot 6\cdot 6 = 6^3 = 216\\\\6^x = 6^3 = 216\\\\\\\boxed{\bf{x=3}}$

8 0

3 years ago

And English class has read the first 384 pages of your novel. If this is 75% of the entire book, how many total pages are in the

Doss [256]

384 pages (0.75) = 288 pages
Add this to the original amount of pages.

384 + 288 = 672 total pages.

3 0

3 years ago

Read 2 more answers

What is the area of this.

Ratling [72]

8x16=? 6x12=? Then add both answers ?+?=?

8 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

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

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

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

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

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

$=5x^4y$

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

2 years ago