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3 years ago

Total distance divided by total time is called what?

1 answer:

azamat3 years ago

6 0

Speed is computed by dividing total distance by total time.

Speed = Total Distance / total time

From the above formula, we can derive the formula of time and distance.

Time = Distance / Speed

Distance = Speed x time

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Select two ratios that are equivalent to 2 : 9.<br>

stira [4]

Answer:

4:18 6:27 plenty more but you said to so

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2 years ago

Ashley, Nikki, Emily, and Brooke are having a contest collecting pennies.

Inessa05 [86]

Ashly then Nikki then Emily

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3 years ago

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A manufacturer of shipping boxes has a box

MissTica

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2 years ago

Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

4 0

2 years ago

2m=-6n-5 literal equation

Bingel [31]

Answer:

read below

Step-by-step explanation:

Alright, archtan /

tan

−

(

)

is the inverse of tangent. Tan is

sin

cos

. Like the inverse of sin, the inverse of tan is also restricted to quadrants 1 and 4.

Knowing this we are solving for the inverse of tan -1. We are basically being asked the question what angle/radian does tan(-1) equal. Using the unit circle we can see that tan(1)= pi/4.

Since the "Odds and Evens Identity" states that tan(-x) = -tan(x). Tan(-1)= -pi/4.

Knowing that tan is negative in quadrants 2 and 4. the answer is in either of those two quadrants. BUT!!! since inverse of tan is restricted to quadrants 1 and 4 we are left with the only answer -pi/4.

the answer is m=-3n-5/2 or n=2m+5/6

8 0

2 years ago