All categories

3 years ago

A line is the locus of points that:

Mathematics

2 answers:

denis23 [38]3 years ago

7 0

A line is the locus of points that: are the same distance from two points. That's the equation for a line.

If helped mark me the brainiest!!

Nataly_w [17]3 years ago

5 0

Answer is b Bc the total lof dermetion will be correct

You might be interested in

What is the slope of (-8, 2) and (1, -4).

Airida [17]

Answer:

-2/3

Step-by-step explanation:

To find the slope of two points

m = (y2-y1)/ (x2-x1)

= (-4-2)/(1--8)

= (-4-2)/(1+8)

= -6/9

-2/3

8 0

4 years ago

Read 2 more answers

To complete a task in 15 days a company needs 4 people each working for 8 hours per day. the company decides to have 5 people ea

Lilit [14]

It would take 16 days
explanation
if it takes 15 for 4 people to work 8 hours to complete the task, it would be a total of 480 hours. I multiplied 15x4x8. i multiplied 5x6 to get 30. I have to find the third value so i divided 30 from 480 to get 16.

3 0

4 years ago

Question 8 8 pts + 4 pts for the best answer Thanks! Question is attached.

elena-14-01-66 [18.8K]

The answer will be -3

5 0

3 years ago

Read 2 more answers

A bar graph titled, Bubbles blown by each gum chewer, plots Number of bubbles, versus Person, for 4 people; Manueala, Sophia, Ja

emmainna [20.7K]

Answer:

Step-by-step explanation:

Mean x Number of gum chewers = Total bubbles

4x4=16

add us the bubbles we do know

4+5+6=15

16-15=1

Hannah blew 1

6 0

4 years ago

Read 2 more answers

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

4 years ago