All categories

2 years ago

William is in the middle of swimming in a ten-mile race. He swim

Mathematics

2 answers:

makvit [3.9K]2 years ago

8 0

Answer:

I made a graph for you, hope it helps.

fiasKO [112]2 years ago

5 0

Answer:

option D is correct

Step-by-step explanation:

i took the test

You might be interested in

Find the slope of the given line, if it is defined.

il63 [147K]

Answer: A. The slope is $-\dfrac{4}{3}$ .

Step-by-step explanation:

The slope of a line helps to specify the direction of the line.

The slope of a line that passes through (a,b) and (c,d) is given by :-

$\text{Slope}=\dfrac{d-b}{c-a}$

Then the slope of a line passes through origin i.e. (0,0) and (-9,12) is given by :-

$\text{Slope}=\dfrac{12-0}{-9-0}\\\\\Rightarrow\ \text{Slope}=\dfrac{12}{-9}\\\\\Rightarrow\ \text{Slope}=-\dfrac{4}{3}$

Hence, the slope is $-\dfrac{4}{3}$ .

8 0

2 years ago

Long-Term Mehlum

pentagon [3]

Answer:

b b.b b b b b b b b b b b b b b b b b b

Step-by-step explanation:

b b b

3 0

3 years ago

Read 2 more answers

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

2 years ago

Read 2 more answers

Where is 27/5 on a number line

Stolb23 [73]

Between 5 and 6 - 5.4.

8 0

2 years ago

Read 2 more answers

I have 8 pennies and some quarters, nickels, and dimes. The number of pennies plus nickels equals the number of dimes. I have 3

GenaCL600 [577]

The answer is D. 10 nickels

3 0

2 years ago

Read 2 more answers