HELP ME OUT PLEASE!!!

Jackie earns $172 per week at her part time job. She is saving this money to buy a used car that cost $2000. At this rate, how m

vodomira [7]

Answer:

12 weeks

Step-by-step explanation:

It will take Jackie 12 weeks to buy her car because she earns 172 per week.

If you divide 2000 buy 172 you end up with 11.6

if you multiply 11 and 172 you get 1,892 which is just a little under 2000

if you add another week = 172$ you will arrive at 2064 which is enough to buy

her car.

Hope this helps.

3 0

2 years ago

20 points brainlist thanks

Fed [463]

Answer:

The answer Is 10.20

Step-by-step explanation:

d = \/(0-(-2))² + (-6-4)²

d = \/(0+2)² + (-10)²

d = \/(2² + 100)

d = \/(4+100)

d = \/104

d ~ 10,20

4 0

3 years ago

Read 2 more answers

Hello, Can someone please explain on how to do this? thank you!

Marrrta [24]

Steps:

1) determine the domain

2) determine the extreme limits of the function

3) determine critical points (where the derivative is zero)

4) determine the intercepts with the axis

5) do a table

6) put the data on a system of coordinates

7) graph: join the points with the best smooth curve

Solution:

1) domain

The logarithmic function is defined for positive real numbers, then you need to state x - 3 > 0

=> x > 3 <-------- domain

2) extreme limits of the function

Limit log (x - 3) when x → ∞ = ∞

Limit log (x - 3) when x → 3+ = - ∞ => the line x = 3 is a vertical asymptote

3) critical points

dy / dx = 0 => 1 / x - 3 which is never true, so there are not critical points (not relative maxima or minima)

4) determine the intercepts with the axis

x-intercept: y = 0 => log (x - 3) = 0 => x - 3 = 1 => x = 4

y-intercept: The function never intercepts the y-axis because x cannot not be 0.

5) do a table

x                          y = log (x - 3)

limit x → 3+            - ∞

3.000000001        log (3.000000001 -3) = -9

3.0001    log (3.0001 - 3) = - 4

3.1                       log (3.1 - 3) = - 1

4                          log (4 - 3) = 0

13                       log (13 - 3) = 1

103                     log (103 - 3) = 10

lim x → ∞             ∞

Now, with all that information you can graph the function: put the data on the coordinate system and join the points with a smooth curve.

7 0

3 years ago

Read 2 more answers

Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices

Airida [17]

Answer:

25/2

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

$\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt$

Where P, Q are scalar functions

We want to compute

$\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy$

Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.

a) Directly

Let us break down C into 4 paths $\large C_1,C_2,C_3,C_4$ which represents the sides of the rectangle.

$\large C_1$ is the line segment from (0,0) to (5,0)

$\large C_2$ is the line segment from (5,0) to (5,1)

$\large C_3$ is the line segment from (5,1) to (0,1)

$\large C_4$ is the line segment from (0,1) to (0,0)

Then

$\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}$

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize $\large C_1$ as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

$\large \displaystyle\int_{C_1}xydx+x^2dy=0$

Now the second integral. We parametrize $\large C_2$ as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

$\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25$

The third integral. We parametrize $\large C_3$ as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

$\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2$

The fourth integral. We parametrize $\large C_4$ as

r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

$\large \displaystyle\int_{C_4}xydx+x^2dy=0$

So

$\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2$

Now, let us compute the value using Green's theorem.

According with this theorem

$\large \displaystyle\int_{C}Pdx+Qdy=\displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx$

where A is the interior of the rectangle.

so A={(x,y) | 0≤ x≤ 5, 0≤ y≤ 1}

We have

$\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x$

so

$\large \displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx=\displaystyle\int_{0}^{5}\displaystyle\int_{0}^{1}xdydx=\displaystyle\int_{0}^{5}xdx\displaystyle\int_{0}^{1}dy=25/2$

3 0

3 years ago

Construct and interpret a 95% confidence interval to estimate the proportion of the earth that is water. Show all work. Make sur

blondinia [14]

the answer would be 2

3 0

2 years ago