Please help it's definitely not E and i'm pretty sure it's not C

Help me with this problem please right away

svlad2 [7]

Answer:

$6\frac{1}{2} = A$

Step-by-step explanation:

The first step is to use the Distance Formula to figure out the lengths of the two legs:

$\sqrt{[-x_1 + x_2]^{2} + [-y_1 + y_2]^{2}} = D$

R(1, 1) and P(−1, −2) ↷

$\sqrt{[1 + 1]^{2} + [2 + 1]^{2}} = \sqrt{2^{2} + 3^{2}} = \sqrt{4 + 9} = \sqrt{13}\\ \\ \sqrt{13} = RP$

Q(2, −4) and P(−1, −2) ↷

$\sqrt{[1 + 2]^{2} + [2 - 4]^{2}} = \sqrt{[-2]^{2} + 3^{2}} = \sqrt{4 + 9} = \sqrt{13} \\ \\ \sqrt{13} = QP$

* So, by my calculations, we have an isosceles <em>right triangle</em>, and according to the <em>45°-45°-90° triangle theorem</em>, we automatically know that the hypotenuse, <em>RQ</em>, is $\sqrt{26}$ :

30°-60°-90° triangle theorem

2x, x, x√3

↑ ↑ ↑

h leg leg

y

p

o

t

e

n

u

s

e

45°-45°-90° triangle theorem

x√2, x, x

↑ ↑ ↑

h legs

y

p

o

t

e

n

u

s

e

So now, we have to find the area of the triangle by taking half of either height or base, then multiplying that by the height or base, but since this is an isosceles <em>right</em><em> </em><em>triangle</em><em>,</em><em> </em>it does not matter:

$[\frac{h}{2}][b] = A \: \:OR\: \: [\frac{b}{2}][h] = A \: \:OR \: \: \frac{hb}{2} = A$

$[\frac{\sqrt{13}}{2}][\sqrt{13}] = \frac{13}{2} = 6\frac{1}{2} \\ \\ 6\frac{1}{2} = A$

I am joyous to assist you anytime.

8 0

3 years ago

Write the polynomial as a product of a difference and a sum: b^2-4/9

Alenkinab [10]

Answer:

$\sf \huge \boxed{ \boxed{(b + \frac{2}{3} )(b - \frac{2}{3} )}}$

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>

algebra
PEMDAS

<h3>tips and formulas:</h3>

${x}^{2} - {y}^{2} = (x + y)(x - y)$

<h3>let's solve:</h3>

$\sf \: rewrite : \\ ( {b})^{2} - ( \frac{2}{3} {)}^{2}$
$\sf use \: the \: formula : \\ (b + \frac{2}{3} )(b - \frac{2}{3} )$

4 0

3 years ago

D is the midpoint of CE. E has coordinates (−8, −3), and D has coordinates (2, 3). Find the coordinates of C.

zvonat [6]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ C(\stackrel{x_1}{x}~,~\stackrel{y_1}{y})\qquad E(\stackrel{x_2}{-8}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-8+x}{2}~~,~~\cfrac{-3+y}{2} \right)=\stackrel{\stackrel{midpoint}{D}}{(2,3)}\implies \begin{cases} \cfrac{-8+x}{2}=2\\[1em] -8+x=4\\ \boxed{x=12}\\[-0.5em] \hrulefill\\ \cfrac{-3+y}{2}=3\\[1em] -3+y=6\\ \boxed{y=9} \end{cases}$

8 0

3 years ago

This is due tonight pls

Anettt [7]

Answer:

I think it's A.3

Step-by-step explanation:

if you didvide the 7.8 by 2.6 then you will get 3, therfore you will have to multiply the original (2.6) times 3 in order to get the scaled image (7.8).

7 0

3 years ago

Read 2 more answers

GEOMETRY!!!! Will give brainliest

nadya68 [22]

Answer: rectangle

Hope itz right

5 0

3 years ago