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3 years ago

12

Help Would Be Greatly Appreciated :)

1 answer:

aliina [53]3 years ago

4 0

Answer:

n is $y=\frac{1}{2}x+b$ , p is $y=\frac{1}{2}x-5$ , and r has infinitely many different lines that have this property.

Step-by-step explanation:

If two lines have no solutions, that means that they are parallel. The slopes of two lines are the same if they are parallel. So n's slope is $\frac{1}{2}$ . The y-intercept can be anything except -5, or +b. If two lines have infinitely many solutions, those two lines are exactly the same. So p is just m. If two lines have one and only one solution, that just means the slopes and y-intercepts are different. Since we don't know anything else about line r, we an't model an equation for it.

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Which phrase represents the algebraic expression 3p+6/7p-9

omeli [17]

Answer:

The answer is the sum of three times a number and six, divided by the difference of seven times the number and nine

Step-by-step explanation:

3p = three x a number

7p = seven x a number

3p+6 = sum of three x a number plus six

7p-9 = difference of seven x the number minus nine

(3p+6)/(7p-9) = sum of three times x number plus six, divided by the difference of seven x the number - nine

7 0

3 years ago

HELP HELP HELP HELP PLEASE

stepladder [879]

About 348 hours. Hope this helps you out!

3 0

3 years ago

What is the outlier of the data set 7, 15, 15, 15, 13, 7, 9, 11, 24

eduard

Answer:

24

Step-by-step explanation:

The outlier is the value that is far from the other values in the set

24 is far from the other values

7 0

3 years ago

Help, ill give brainliest

larisa [96]

Answer:

Step-by-step explanation:

a b c

d e f

solution is b,e

5 0

2 years ago

Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)

Triss [41]

$g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}$

The surface element is

$\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

and the integral is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)$

###

To compute the last integral, you can integrate by parts:

$u=r\implies\mathrm du=\mathrm dr$

$\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}$

$\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr$

For this integral, consider a substitution of

$r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds$

$\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds$

$\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds$

$=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds$

and the result above follows.

4 0

3 years ago