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3 years ago

11

Evaluate the equation y = abx when a = 2, b = 1/2, and x = 3. Make into a fraction.

1 answer:

topjm [15]3 years ago

4 0

Abx=2*1/2*3=6/2=3

Hope this helps!

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A regular octagonal pyramid has a base edge of 3m and a lateral area of 60m^2. Find its slant height.

Olin [163]

Octagon = 8-side polygon

<u>Find area of 1 triangle:</u>
Area of one side triangle = 60 ÷ 8 = 7.5 m²

<u>Given area and length, find height:</u>
Area of each triangle = 7.6 m²
Length of each triangle = 3m

Area = 1/2 x base x height
7.5 = 1/2 x 3 x height
Heigth = 5m

Answer: Slanted Height = 5 m

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2 years ago

A study finds a positive correlation between the number of traffic lights on the most-used route between two destinations and th

ICE Princess25 [194]

The correlation is most likely a causation.

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3 years ago

Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt

polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

$\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt$

$=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k$

The first integral is trivial since $(\tan t)'=\sec^2t$ .

The second can be done by substituting $u=t^2-1$ :

$u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C$

The third can be found by integrating by parts:

$u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t$

$\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8$

$\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C$

8 0

3 years ago

What is the area of a sector with a central angle of 131 degrees and a radius of 7.2 ft?

svetoff [14.1K]

Answer:

162.78

Step-by-step explanation:

Use your π =3.14

3.14 times 7.2 times 7.2

5 0

3 years ago

Given: 1; -5; -13 ; -23 ; ...<br><br> Derive a formula for the nth term in the pattern.

mylen [45]

Answer:

f(n) = -n^2 -3n +5

Step-by-step explanation:

Suppose the formula is ...

f(n) = an^2 +bn +c

Then we have ...

f(1) = 1 = a(1^2) +b(1) +c

f(2) = -5 = a(2^2) +b(2) +c

f(3) = -13 = a(3^2) +b(3) +c

__

Here's a way to solve these equations.

Subtract the first equation from the second:

-6 = 3a +b . . . . . 4th equation

Subtract the second equation from the third:

-8 = 5a +b . . . . . 5th equation

Subtract the fourth equation from the fifth:

-2 = 2a

a = -1

Then substituting into the 4th equation to find b, we have ...

-6 = 3(-1) +b

-3 = b

and ...

1 = -1 +(-3) +c . . . . . substituting "a" and "b" into the first equation

5 = c

The formula is ...

f(n) = -n^2 -3n +5

3 0

3 years ago