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3 years ago

Which is an exponential decay function f (x)=3/4 (7/4)x f (x)=2/3 (4/5)-x. F (x)=3/2 (8/7)-x f (x)=1/3 (9/2)x

2 answers:

7 0

If all the x's are exponents, then 3/2 (8/7)^-x is the only one that decays as x gets bigger. That's because the base is greater than 1 and the x is negative.

zysi [14]3 years ago

6 0

To answer your question the answer is c)3/2(8/7)^x

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Mariah spent $9.50 on 9 pounds of limes and pears. Limes cost $0.50 per pound and pears cost $1.50 per pound. Let l be the numbe

madam [21]

L = limes, P = pears

L + P = 9
0.50L + 1.50P = 9.50

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4 years ago

A packet contains 24 pens.

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Alice would receive 15 pens while Henry would receive 9.

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3 years ago

Read 2 more answers

If c is the line segment connecting (x1,y1) to (x2,y2), show that the line integral of xdy-ydx=x1y2-x2y1......this is in a chapt

MatroZZZ [7]

Green's theorem doesn't really apply here. GT relates the line integral over some *closed* connected contour that bounds some region (like a circular path that serves as the boundary to a disk). A line segment doesn't form a region since it's completely one-dimensional.

At any rate, we can still compute the line integral just fine. It's just that GT is irrelevant.

We parameterize the line segment by

$\mathbf r(t)=\langle x_1,y_1\rangle(1-t)+\langle x_2,y_2\rangle t$
$\implies\mathbf r(t)=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle$

with $0\le t\le1$ . Then we find the differential:

$\mathbf r(t)\equiv\langle x,y\rangle=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle$
$\implies\mathrm d\mathbf r\equiv\langle\mathrm dx,\mathrm dy\rangle=\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt$

with $0\le t\le1$ .

Here, the line integral is

$\displaystyle\int_{\mathcal C}x\,\mathrm dy-y\,\mathrm dx=\int_{\mathcal C}\langle-y,x\rangle\cdot\langle\mathrm dx,\mathrm dy\rangle$
$=\displaystyle\int_{t=0}^{t=1}\langle-y_1-(y_2-y_1)t,x_1+(x_2-x_1)t\rangle\cdot\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt$
$=\displaystyle\int_{t=0}^{t=1}(x_1y_2-x_2y_1)\,\mathrm dt$
$=(x_1y_2-x_2y_1)\displaystyle\int_{t=0}^{t=1}\,\mathrm dt$
$=x_1y_2-x_2y_1$

as required.

4 0

4 years ago

Ajar contains 4 red marbles numbered 1 to 4 and 10 blue marbles numbered 1 to 10. A marble is

Reptile [31]

Answer:

a)2/7

b)1/2

c)9/14

d)6/7

Step-by-step explanation:

The jar contains 4 red marbles, numbered 1 to 4 which means

Red marbles = (R1) , (R2) , (R3) , (R4)

It also contains 10 blue marbles numbered 1 to 10 which means

Blue marbles = (B1) , (B2) , (B3) , (B4) , (B5) , (B6) , (B7) , (B8) , (B9) , (B10) .

We can calculate total marbles = 4red +10 blues

=14marbled

Therefore, total marbles= 14

The marbles that has even number = (R2) , (R4) ,(B2) , (B4) , (B6) , (B8) , (B10) =7

Total number of Blue marbles = 10

Blue and even marbles = 5

(a) The marble is red

P(The marble is red)=total number of red marbles/Total number of marbles

=4/14

=2/7

(b) The marble is odd-numbered

Blue marbles with odd number= (B1) , (B3) , (B5) , (B7) , (B9) ,

Red marbles with odd number = (R1) , (R3)

Number of odd numbered =(5+2)=7

P(marble is odd-numbered )= Number of odd numbered/ Total number of marbles

P(marble is odd-numbered )=7/14

=1/2

Total number of red marbles = 14

Number of red and odd marbles = 2

The marbles that has odd number = (R1) , (R3) ,(B1) , (B3) , (B5) , (B7) , (B9) =7

n(red or even )= n(red) + n(odd)- n(red and odd)

=4+7-2

P(red or odd numbered)= (number of red or odd)/(total number of the marble)

= 9/14

(d) The marble is blue or even-numbered?

Number of Blue and even marbles = 5

Total number of Blue marbles = 10

Number of blue that are even= 5

The marbles that has even number = (R2) , (R4) ,(B2) , (B4) , (B6) , (B8) , (B10)

n(Blue or even )= n(Blue) + n(even)- n(Blue and even)

= 10+7-5 =12

Now , the probability the marble is blue or even numbered can be calculated as

P(blue or even numbered)= (number of Blue or even)/(total number of the marble)

= 12/14

= 6/7

6 0

3 years ago

Find the product of the functions: (fxg)(x). f(x)=7x 8, g(x)=5x 2

IgorC [24]

All you have to do is plug in the #'s. or equation to eachother to solve the problem

8 0

3 years ago