PLEASE HELP ASAP!!! 100 POINTS!!!

Find the length of a diagonal of a square with sides 10in. long

Ghella [55]

Hey!!!
Answer is....
diagonals of square bisect each other at right angles.
let diagonal be 2x
Using Pythagoras in a triangle enclosed by a side and 2 half diagonals...
${x}^{2} + {x}^{2} = {10}^{2}$
$2 {x}^{2} = 100$
${x }^{2} = 50$
$x = \sqrt{50}$
side of a diagonal=2 sqrt50
$= 10 \sqrt{2}$
cm......

HOPE IT HELPS YOU '_'

3 0

2 years ago

Read 2 more answers

A line passes through the points (-1, -6) and (-4, 3). What is the equation of this line in slope-intercept form?

g100num [7]

Since we have two ordered pairs, we can find the slope by plugging them into (y2-y1)/(x2-x1) which translates to (3-(-6))/(-4-(-1))=9/-3=-3

Since the slope is -3, we have the equation y=-3x+b, but we don't know b. So we plug in values for x and y and solve for b.

-6=-3(-1)+b

-6=3+b

-9=b

So the y-intercept is -9 and the full equation is y=-3x-9 which means A is the only correct answer.

Hope this helped!

4 0

3 years ago

HEEELP

Drupady [299]

Answer: (c)

Step-by-step explanation:

Given

$f(x)=\dfrac{1}{x-3}\\\\g(x)=\sqrt{x+5}$

Here, $\sqrt{x+5}\ \text{is always greater than equal to 0}\\\Rightarrow x+5\geq 0\\\Rightarrow x\geq -5\quad \ldots(i)$

To get $f\left(g(x)\right)$ , replace $x$ in $f(x)$ by $g(x)\ \text{i.e. by}\ \sqrt{x+5}$

$\Rightarrow f\left(g(x)\right)=\dfrac{1}{\sqrt{x+5}-3}\\\\\text{Denominator must not be equal to 0}\\\\\therefore \sqrt{x+5}-3\neq0\\\Rightarrow \sqrt{x+5}\neq 3\\\Rightarrow x+5\neq 9\\\Rightarrow x\neq 4\quad \ldots(ii)$