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3 years ago

5

Kerry has a large triangular price of fabric that she wants to attach to the ceiling in her bedroom. The sides of the piece of f

abric measure 4.8 ft, 6.4 ft, and 8 ft. Is the fabric in the shape of a right triangle? Explain.

1 answer:

kotykmax [81]3 years ago

7 0

An easy way to test this is with the equation $A^{2}+B ^{2}=C ^{2}$ this means that if you add 4.8²+6.4² does it equal 8²? if so then it is a right triangle. This is called the<span> Pythagorean theorem. So the answer is, yes it is a right triangle. </span>

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PLEASE HELP ITS DUE IN A FEW MINUTES

denis23 [38]

1. a+3a+4b=5b
2. a+3a=5b-4b
3. 4a=5b-4b
4. 4a=b
a=1/4b
(I think 1/4b is the answer) hope this helps :)

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3 years ago

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Find the midpoint of the segment with the following endpoints.<br> (2,9) and (8,1)

vazorg [7]

Answer:

$\displaystyle (5,5)$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Algebra I</u>

Coordinates (x, y)
Midpoint Formula: $\displaystyle (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Step-by-step explanation:

<u>Step 1: Define</u>

Point (2, 9)

Point (8, 1)

<u>Step 2: Identify</u>

(2, 9) → x₁ = 2, y₁ = 9

(8, 1) → x₂ = 8, y₂ = 1

<u>Step 3: Find Midpoint</u>

Simply plug in your coordinates into the midpoint formula to find midpoint

Substitute in points [Midpoint Formula]: $\displaystyle (\frac{2+8}{2},\frac{9+1}{2})$
[Fractions] Add: $\displaystyle (\frac{10}{2},\frac{10}{2})$
[Fractions] Divide: $\displaystyle (5,5)$

7 0

3 years ago

Lerato receive an increase of 5% on her current salary of R12500 per month. What will her new salary be after the increase

Romashka [77]

$12,500 x 5% = $625.
$12,500 + $625 = $13,125.

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3 years ago

I just need 5 please help show work and I’ll give brainliest

Ostrovityanka [42]

Hello from MrBillDoesMath!

Answer:

See Discussion below

Discussion:

A function f is even if f(-x) = f(x)

f(x) f(-x) Are they equal?

----------------------------------------------------------------------------------------

-x^8 + 2x^6-5x -(-x)^8 + 2(-x)^6 + 5x No

3 abs(x) - 4 3 abs(-x) -4 Yes

log5 x^2 log5 (-x)^2 Yes

(6x)^ (1/7) (-6x)^(1/7) No

e^(x^2-x) e^( (-x)^2+x) No

(x^8 +5x^2)^(-1) ( (-x)^8 + 5 (-x)^2) ^(-1) Yes

Answers with Yes, above are even functions.

Regards,

MrB

P.S. I'll be on vacation from Friday, Dec 22 to Jan 2, 2019. Have a Great New Year!

6 0

3 years ago

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

3 years ago

Read 2 more answers