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3 years ago

I’m going to continue using up all my points

Mathematics

2 answers:

Sveta_85 [38]3 years ago

6 0

Answer:

Black

Step-by-step explanation:

I like black because it's a color that looks good on anything.

Hope this helped!

lakkis [162]3 years ago

5 0

Answer:

My favorite color is purple (any shade of purple) My favorite animal is between birds and dogs. Have an amazing day!

You might be interested in

Consider the following. C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1), starting at (0, 0)

horsena [70]

Answer:

$\mathbf{r_1 = (t,0) \implies t = 0 \ to \ 1}$

$\mathbf{r_2 = (2-t,t-1) \implies t = 1 \ to \ 2}$

$\mathbf{r_3 = (0,3-t) \implies t = 2 \ to \ 3}$

$\mathbf{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+11}{6}}$

Step-by-step explanation:

Given that:

C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1), starting at (0, 0)

a. Find a piecewise smooth parametrization of the path C.

r(t) = { 0

If C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1),

Then:

$C_1 = (0,0) \\ \\ C_2 = (1,0) \\ \\ C_3 = (0,1)$

Also:

$\mathtt{r_1 = (0,0) + t(1,0) = (t,0) }$

$\mathbf{r_1 = (t,0) \implies t = 0 \ to \ 1}$

$\mathtt{r_2 = (1,0) + t(-1,1) = (1- t,t) }$

$\mathbf{r_2 = (2-t,t-1) \implies t = 1 \ to \ 2}$

$\mathtt{r_3 = (0,1) + t(0,-1) = (0,1-t) }$

$\mathbf{r_3 = (0,3-t) \implies t = 2 \ to \ 3}$

b Evaluate :

Integral of (x+2y^1/2)ds

$\mathtt{\int \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \int \limits ^1_{0} \ (t + 0) \sqrt{1} } \\ \\ \mathtt{ \int \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \begin {pmatrix} \dfrac{t^2}{2} \end {pmatrix} }^1_0 \\ \\ \mathtt{\int \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \dfrac{1}{2}}$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \int \limits (x+2 \sqrt{y} \sqrt{(\dfrac{dx}{dt})^2 + (\dfrac{dy}{dt})^2 \ dt } }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \int \limits 2- t + 2\sqrt{t-1} \ \sqrt{1+1} }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} \int \limits^2_1 2- t + 2\sqrt{t-1} \ dt }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} 2t - \dfrac{t^2}{2}+ \dfrac{2(t-1)^{3/2}}{3} (2) \end {pmatrix} ^2_1}$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} 2 -\dfrac{1}{2} (4-1)+\dfrac{4}{3} (1)^{3/2} -0 \end {pmatrix} }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} 2 -\dfrac{3}{2} + \dfrac{4}{3} \end {pmatrix} }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} \dfrac{12-9+8}{6} \end {pmatrix} }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} \dfrac{11}{6} \end {pmatrix} }$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \dfrac{ \sqrt{2} }{6} \ (11 )}$

$\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \dfrac{ 11 \sqrt{2} }{6}}$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \int \limits ^3_2 0+2 \sqrt{3-t} \ \sqrt{0+1} }$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \int \limits ^3_2 2 \sqrt{3-t} \ dt}$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \int \limits^3_2 \begin {pmatrix} \dfrac{-2(3-t)^{3/2}}{3} (2) \end {pmatrix}^3_2 }$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = -\dfrac{4}{3} [(0)-(1)]}$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = -\dfrac{4}{3} [-(1)]}$

$\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \dfrac{4}{3}}$

$\mathtt{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}}{6}+\dfrac{1}{2}+ \dfrac{4}{3}}$

$\mathtt{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+3+8}{6}}$

$\mathbf{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+11}{6}}$

5 0

3 years ago

The graph of a sinusoidal function intersects its midline at (0,5) and then has a maximum point at (\pi,6)

lesya692 [45]

First, let's use the given information to determine the function's amplitude, midline, and period.

Then, we should determine whether to use a sine or a cosine function, based on the point where x=0.

Finally, we should determine the parameters of the function's formula by considering all the above.

Determining the amplitude, midline, and period

The midline intersection is at y=5 so this is the midline.

The maximum point is 1 unit above the midline, so the amplitude is 1.

The maximum point is π units to the right of the midline intersection, so the period is 4 * π.

Determining the type of function to use

Since the graph intersects its midline at x=0, we should use thesine function and not the cosine function.

This means there's no horizontal shift, so the function is of the form -

a sin(bx)+d

Since the midline intersection at x=0 is followed by a maximumpoint, we know that a > 0.

The amplitude is 1, so |a| = 1. Since a >0 we can conclude that a=1.

The midline is y=5, so d=5.

The period is 4π so b = 2π / 4π = 1/2 simplified.

$f(x) = 1 sin ( \dfrac{1}{2}x)+5$ <span>= Solution </span>

4 0

4 years ago

Allison worked 48 hours this week. If she makes $15.00 per hour, how much extra money did she make on her overtime hours?

GuDViN [60]

Allison worked 40 regular hours and 8 overtime hours.

8*15*1.5=180

6 0

3 years ago

Please help! I have no idea how to do this, it’s very confusing

Vsevolod [243]

I like that you left How you did it on there. When I was in school I did it the same way. I would assume you’re using is over of equals percent over 100. The problem however is that the above number is always the total which means it should be on the bottom.

X/72= 45/100. Similar to what you have written. Solve the proportion. 100x=72x45
100x= 3240
X = 32.4

3 0

3 years ago

the larger of two numbers is 1 less than 3 times the smaller number. if twice the smaller is increased by the larger number, the

Y_Kistochka [10]

x = 3y - 1

2y + x = x + 18

2y + 3y - 1 = 3y - 1 + 18

5y - 1 = 3y + 17

2y = 18

y = 9

x = 3(9) - 1

x = 27 - 1

x = 26

5 0

3 years ago