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3 years ago

I need help please and thank you

Mathematics

2 answers:

d1i1m1o1n [39]3 years ago

8 0

Answer:

8.08

Step-by-step explanation:

mylen [45]3 years ago

8 0

8.5%= 0.085
0.085 x $95 = 8.075

$8.08 to nearest 100

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1. 42<br><br> 2.70<br><br> 3.35<br><br> 4.84<br><br> Helpppppppppp

vitfil [10]

Answer = 84

Step-by-step explanation :

Area = b • h

14 = b
6 = h

14 • 6 = 84

4 0

3 years ago

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Triangle Inequality Theorem: The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

Dovator [93]

Answer:

1) yes 2) no 3) yes

Step-by-step explanation:

The lengths of the two shorter sides must be greater than the length of the longest side.

1) 8 + 9 > 10 yes

2) 1 + 1 = 2 no

3) 6 + 8 > 9 yes

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3 years ago

HELP PLEASEE THIS IS MY LAST QUESTION ILL MARK BRAINLIEST

RoseWind [281]

Answer:

F. y = 9.5x + 22.5

Step-by-step explanation:

let the number of shirts be x

For every shirt he pays $9.50 and $22.5 is the additional price

8 0

3 years ago

What is the answer and how to solve it and give me the right answer

Keith_Richards [23]

Answer:

2/10 or in its simplest form 1/5

Step-by-step explanation:

there is 2 grey marbles so 2/10 (there is 10 marbles altogether)

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so the answer is 1/5

7 0

3 years ago

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Find the limit

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

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