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3 years ago

Write a point slope form of the line that passes through (6,1) and it is parallel to a line with a slope of -3

Mathematics

1 answer:

Dafna1 [17]3 years ago

8 0

Answer:

Step-by-step explanation:

y - 1 = -3(x - 6)

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The volume of a solid right pyramid with a square base is v units3 and the length of the base edge is y units. Write the express

Wewaii [24]

Answer:

h = V/y^2

Step-by-step explanation:

The volume of a right pyramid is given by the expression = Area of base × height.

Area of base = Area of square at the base

= y × y = y^2.

Now, we substitute these values:

V = y^2 × height

height = V/y^2

6 0

3 years ago

What is 14x+2 and 2(7x+)?

Svetach [21]

Answer:

#1 is 2

Step-by-step explanation:

sorry i couldnt get both

5 0

3 years ago

finding the distance along a diagonal with the distance formula requires you evaluate a square root. true or false ?

bija089 [108]

Answer:

true

Step-by-step explanation:

Pythagoras is simply c² = a² + b².

so, it does not matter which side of the triangle we need, we will always have to solve a square root for the final result.

7 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

I will mark you brainliest!!

stepan [7]

Answer:

C: .94

Step-by-step explanation:

7e^x=18

x=ln(18/7)

x=ln(2.5714)

x=.94

3 0

3 years ago