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3 years ago

Please please please help me with this question and show your work.

Mathematics

1 answer:

lawyer [7]3 years ago

4 0

The correct answer is D. You can see It is the biggest bar and so you can estimate the correct answer.

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Cleo and jandi are having a hollowed party. They have 3 1/2 pounds of candy to divide into 14 goodie bags. How much candy will b

OLEGan [10]

Answer: 1/4 pound(s) of candy in each bag

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Please help! I need to find the length of a rectangular prisim with the height as 1.5 inches and the width as 2.5 inches for a p

adelina 88 [10]

I guess try 2.0 inches for the length

4 0

4 years ago

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

Read 2 more answers

Sabendo que "K" satisfaz a equação {2(k-8) + 3(-k+1) = -4k +11} Os valores reais de x que satisfazem a equação 15x² - kx + 1 = 0

erastova [34]

Answer:

D) 1/5 e 1/3

Step-by-step explanation:

You have the following quadratic equation:

$15x^2-kx+1=0$ (1)

In order to find the values of x that are solution to the equation (1), you first find the solution for k in the following equation:

$2(k-8)+3(-k+1)=-4k+11\\\\2k-16-3k+3=-4k+11\\\\2k-3k+4k=11+16-3\\\\3k=24\\\\k=8$

Next, you replace the previous value of k in the equation (1) and you use the quadratic formula to find the roots:

$15x^2-8x+1=0\\\\x_{1,2}=\frac{-(-8)\pm \sqrt{(-8)^2-4(15)(1)}}{2(15)}\\\\x_{1,2}=\frac{8\pm 2}{30}\\\\x_1=\frac{1}{5}\\\\x_2=\frac{1}{3}$

Then, the roots of the equation (1) are

D) 1/5 e 1/3

5 0

4 years ago

HELP ME PLEASE I REALLY NEED THE HELP

dedylja [7]

Answer:

Answer = 126 square inches

Step-by-step explanation:

Square area formula is Area = s^2 (replace the s with the side length)

Square area --> Area = 4^2 Area = 16in

Parallelogram area formula is Area = h * b (replace h with height length and replace b with base length)

Parallelogram area --> Area = 3.5 * 10 Area = 35

Rectangle area formula is Area = w * h (replace width and height)

Rectangle area --> Area = 12.5 * 6 Area = 75

Then add them all up 75 + 35 + 16 = 126 square inches

3 0

3 years ago