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MariettaO [177]
3 years ago
5

Calculate the indefinite integral : (assume t > a and t >

Mathematics
1 answer:
ElenaW [278]3 years ago
4 0
So here's your problem: \int\limits {[t(t-a)(t-b)]} \, dt.  The easiest way to do this is to distribute that whole thing out by FOIL-ing to get \int\limits {t^3-t^2b-t^2a+abt} \, dt.  Now the integration is straightforward. We are integrating with respect to t, so treat a and b like "regular" numbers.  Your integration, before simplifying, is \frac{t^4}{4}- \frac{bt^3}{3}- \frac{at^3}{3}+ \frac{abt^2}{2}.  Those 2 terms in the middle have the same denominator and power on the t, so we will combine those as like to get \frac{t^4}{4}- \frac{abt^3}{3}+ \frac{abt^2}{2}+C
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MULTIPLE CHOICE GEOMOTRY QUESTIOON
rosijanka [135]

Answer: D

Step-by-step explanation: Im vv sure, is this a test?

7 0
3 years ago
Could someone explain this to me?
Ilya [14]
This is a simple differentiation problem. Let's start by taking the derivative of both sides (with respect to x):
5y^4\frac{dy}{dx}+6y^2x+6x^2y\frac{dy}{dx}+20x^3 = 0

Simplify:
\frac{dy}{dx} (5y^4+6x^2y) + 6y^2x+20x^3 = 0

Solve for dy/dx:
\frac{dy}{dx} = \frac{-6y^2x-20x^3}{5y^4+6x^2y}

Now, plug in the given points:
\frac{dy}{dx} = \frac{-6(2)^2(-1)-20(-1)^3}{5(2)^4+6(-1)^2(2)} = \frac{24+20}{80+12}

Further simplification gives:
\frac{dy}{dx}|_{(2,-1)} =\frac{44}{92} =  \frac{11}{23}

So, your answer is 11/23 or B.
7 0
4 years ago
Jessie works at least 36 hours a week. She works as a math
Olegator [25]

Answer:

The feasible region is unbounded which is the shaded portion in the graph.

Step-by-step explanation:

Given that, the number of hours Jessie works as a teacher and tutor is y and x respectively, which can't be negative.

So, x\ge0\;\cdots(i)

y\ge0\;\cdots(ii)

She works at least 36 hours a week.

\Rightarrow x+y\ge 36

\Rightarrow y\ge 36-x\;\cdots (iii)

She works as a teacher for al least twice the number of hours she works as a tutor.

\Rightarrow y\ge2x\;\cdots(iv)

The boundary of the inequalities from equations (i), (ii), (iii), and (iv) are

x=0, y=0, y=36-x and y=2x.

The feasible region is the shaded portion which is the unbounded region as shown in the graph. Any point P(x,y) from the feasible region will satisfies all the constraints.

5 0
3 years ago
IMAGE 1: Find the value of the indicated trigonometric ratio cos α
Sphinxa [80]

Answer:

28.07° ; 70.53° ; 53.13° ; 36.87°

Step-by-step explanation:

Using PYTHAGORAS RULE :

IMAGE 1:

value of cos α equals :

cos α = Adjacent / Hypotenus

cos α = 15 / 17

cos α = 0.88235

α = cos^-1 (0.88235)

α = 28.07°

IMAGE 2:

sin β = Opposite / Hypotenus

sin β = 14√2 / 21

sin β = 0.9428

β = sin^-1(0.9428)

β = 70.53°

IMAGE 3:

From trigonometry :

cot = 1 / tan

Tan = Opposite / Adjacent

Therefore, Cot = Adjacent / Opposite

cot γ = 3/4

cot γ = 0.75

γ = cot^-1 (0.75)

γ = 53.13°

IMAGE 4:

tan α = Opposite / Adjacent

tan α = 15 / 20

tan α = 0.75

α = tan^-1 (0.75)

α = 36.87°

6 0
3 years ago
I added the picture of the question.
Alex787 [66]

Answer:

A) 4

Step-by-step explanation:

Area of a triangle formula:

A = 0.5bh

Given:

b = 4

h = 2

Work:

A = 0.5bh

A = 0.5(4)(2)

A = 2(2)

A = 4

4 0
3 years ago
Read 2 more answers
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