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3 years ago

14

Jamal is running 42 kilometers. The amount of time t it takes Jamal to run varies inversely with the rate, r, at which he runs,

write the inverse variation equation that represents this situation

1 answer:

leva [86]3 years ago

6 0

Answer:

t α 1 /r ; t = k/r

Step-by-step explanation:

Given that :

Time taken, t varies inverse as rate, r

MATHEMATICALLY,

t α 1 /r

Hence,

t = k/r ;

Where, k is the proportionality constant

Hence, the situation could be expressed as :

t = k/r OR

t α 1 /r

You might be interested in

How do you do (a) and (b)?

bulgar [2K]

Answer:

See solution below

Step-by-step explanation:

(a) If n=0 or 1, the equation

(1) $y' = a(t)y + f(t)y^n$

would be a simple linear differential equation. So, we can assume that n is different to 0 or 1.

Let's use the following substitution:

(2) $z=y^{n-1}$

Taking the derivative implicitly and using the chain rule:

(3) $z'=(1-n)y^{-n}y'$

Multiplying equation (1) on both sides by

$(1-n)y^{-n}$

we obtain the equation

$(1-n)y^{-n}y' = (1-n)y^{-n}a(t)y+(1-n)y^{-n}f(t)y^n$

reordering:

$(1-n)y^{-n}y' = (1-n)y^{-n}ya(t)+(1-n)y^{-n}y^nf(t)$

$(1-n)y^{-n}y' = (1-n)y^{1-n}a(t)+(1-n)y^{0}f(t)$

$(1-n)y^{-n}y' = (1-n)y^{1-n}a(t)+(1-n)f(t)$

Now, using (2) and (3) we get:

$z'= (1-n)za(t)+(1-n)f(t)$

which is an ordinary linear differential equation with unknown function z(t).

(b)

The equation we want to solve is

(4) $xy'+ y = x^4 y^3$

Here, our independent variable is x (instead of t)

Assuming x different to 0, we divide both sides by x to obtain:

$y'+\frac{1}{x}y = x^3 y^3$

$y' = -\frac{1}{x}y+x^3 y^3$

Which is an equation of the form (1) with

$a(x)=-\frac{1}{x}$

$f(x)=x^3$

$n=3$

So, if we substitute

$z=y^{-2}$

we transform equation (4) in the lineal equation

(5) $z'=\frac{2}{x}z-2x^3$

and this is an ordinary lineal differential equation of first order whose

integrating factor is

$e^{\int (-\frac{2}{x})dx}$

but

$e^{\int (-\frac{2}{x})dx}=e^{-2\int \frac{dx}{x}}=e^{-2ln(x)}=e^{ln(x^{-2})}=x^{-2}=\frac{1}{x^2}$

Similarly,

$e^{\int (\frac{2}{x})dx}=x^2$

and the general solution of (5) is then

$z(x)=x^2\int (\frac{-2x^3}{x^2})dx+Cx^2=-2x^2\int xdx+Cx^2=\\\ -2x^2\frac{x^2}{2}+Cx^2=-x^4+Cx^2$

where C is any real constant

Reversing the substitution

$z=y^{-2}$

we obtain the general solution of (4)

$y=\sqrt{\frac{1}{z}}=\sqrt{\frac{1}{-x^4+Cx^2}}$

Attached there is a sketch of several particular solutions corresponding to C=1,4,6

It is worth noticing that the solutions are not defined on x=0 and for C<0

4 0

3 years ago

In 1992, there were 41 comma 984 shopping centers in a certain country. In 2002, there were 48 comma 225. (a) Write an equati

Illusion [34]

Answer:

Step-by-step explanation:

The equation expressing the number y of shopping centers in terms of the number x of years after 1992 will be LINEAR in nature

The standard equation is expressed as;

y = mx+c

m is the slope

c is the intercept

If in 1992, there were 41,984 shopping centers in a certain country and in 2002, there were 48,225 we can express this as coordinate points

(0, 41984) and (10, 48,225)

Get the slope

m = 48,225-41,984/10-0

m = 6241/10

m = 624.1

Get the intercept, substitute m = 624.1 and the point (0,48225) into the formula y = mx+c

48225 = 624.1(0)+c

48225 = 0+c

c = 48225

Get the required equation

y = 624.1x+48225

5 0

3 years ago

Help me <br> 3.9x0.5+4 5/6 / 3 3/7

sergey [27]

3.9 · 0.5 + 4 5/6 ÷ 3 3/7
1.95 + 4 5/6 ÷ 3 3/7 multiplication left ot right
1.95 + 29/6 ÷ 24/7 convert to improper fraction
1.95 + 29/6 · 7/24 divide by fraction same as multiplication by reciprocal
1.95 + 203/144 multiply
1.95 + 1.40 (and a teensy bit more)
3.35

5 0

3 years ago

If you have 60 and get 45 as discount how much the percentage

WINSTONCH [101]

Let x% be the percentage
60×(1-x%)=45
1-x%=0.75
x=25%

7 0

3 years ago

You invested $23,000 in two accounts paying 6% and 7% annual interest, respectively. If the total interest earned for the year w

Sedbober [7]

Answer:

$ 18,000 was invested at 7%, and $ 5,000 was invested at 6%.

Step-by-step explanation:

Given that I invested $ 23,000 in two accounts paying 6% and 7% annual interest, respectively, and the total interest earned for the year was $ 1,560, to determine how much was invested at each rate, the following calculation must be performed:

23,000 x 0.07 + 0 x 0.06 = 1,610

22,000 x 0.07 x 1,000 x 0.06 = 1,600

21,000 x 0.07 x 2,000 x 0.06 = 1,590

20,000 x 0.07 x 3,000 x 0.06 = 1,580

19,000 x 0.07 x 4,000 x 0.06 = 1,570

18,000 x 0.07 x 5,000 x 0.06 = 1,560

Thus, $ 18,000 was invested at 7%, and $ 5,000 was invested at 6%.

3 0

3 years ago