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Deffense [45]
3 years ago
8

When Harvey picked up his rental compact car, the odometer read 44,341 miles. When he returned it 4 days later, it read 45,230.

What was his rental on the car if the daily rate was $28.99 and $0.30 per mile with the first 50 miles being free?
Mathematics
2 answers:
Georgia [21]3 years ago
8 0

Answer: daily rate = $28.99

= 28.99*4days= 115.96

Given the 50 miles being free

=0.30*50 = 15.00

= 15-115.96= 100.96

= 100.96

Step-by-step explanation:

dlinn [17]3 years ago
8 0

Answer:

$376.66

Step-by-step explanation:

$28.99 per day for 4 days = $115.96

45,230 miles - 44,341 miles = 889 miles

First 50 miles are free so he owes for 839 miles

The cost per mile is $0.30 so 839($0.30) = $251.70

$251.70 + $115.96 = $376.66

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Find the absolute maximum and minimum values of f(x, y) = x+y+ p 1 − x 2 − y 2 on the quarter disc {(x, y) | x ≥ 0, y ≥ 0, x2 +
Andreas93 [3]

Answer:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

Step-by-step explanation:

In order to find the absolute max and min, we need to analyse the region inside the quarter disc and the region at the limit of the disc:

<u>Region inside the quarter disc:</u>

There could be Minimums and Maximums, if:

∇f(x,y)=(0,0) (gradient)

we develop:

(1-2x, 1-2y)=(0,0)

x=1/2

y=1/2

Critic point P(1/2,1/2) is inside the quarter disc.

f(P)=1/2+1/2+p1-1/4-1/4=1/2+p1

f(0,0)=p1

We see that:

f(P)>f(0,0), then P(1/2,1/2) is a maximum relative

<u>Region at the limit of the disc:</u>

We use the Method of Lagrange Multipliers, when we need to find a max o min from a f(x,y) subject to a constraint g(x,y); g(x,y)=K (constant). In our case the constraint are the curves of the quarter disc:

g1(x, y)=x^2+y^2=1

g2(x, y)=x=0

g3(x, y)=y=0

We can obtain the critical points (maximums and minimums) subject to the constraint by solving the system of equations:

∇f(x,y)=λ∇g(x,y) ; (gradient)

g(x,y)=K

<u>Analyse in g2:</u>

x=0;

1-2y=0;

y=1/2

Q(0,1/2) critical point

f(Q)=1/4+p1

We do the same reflexion as for P. Q is a maximum relative

<u>Analyse in g3:</u>

y=0;

1-2x=0;

x=1/2

R(1/2,0) critical point

f(R)=1/4+p1

We do the same reflexion as for P. R is a maximum relative

<u>Analyse in g1:</u>

(1-2x, 1-2y)=λ(2x,2y)

x^2+y^2=1

Developing:

x=1/(2λ+2)

y=1/(2λ+2)

x^2+y^2=1

So:

(1/(2λ+2))^2+(1/(2λ+2))^2=1

\lambda_{1}=\sqrt{1/2}*-1 =-0.29

\lambda_{2}=-\sqrt{1/2}*-1 =-1.71

\lambda_{2} give us (x,y) values negatives, outside the region, so we do not take it in account

For \lambda_{1}: S(x,y)=(0.70, 070)

and

f(S)=0.70+0.70+p1-0.70^2-0.70^2=0.42+p1

We do the same reflexion as for P. S is a maximum relative

<u>Points limits between g1, g2 y g3</u>

we need also to analyse the points limits between g1, g2 y g3, that means U(0,0), V(1,0), W(0,1)

f(U)=p1

f(V)=p1

f(W)=p1

We can see that this 3 points are minimums relatives.

<u>Conclusion:</u>

We compare all the critical points P,Q,R,S,T,U,V,W an their respective values f(x,y). We find that:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

4 0
3 years ago
Need help ASAP! (Geometry 1)
melisa1 [442]

Answer:

The Length of third side is 52 units.

Step-by-step explanation:

Given:

Length of First side = 48

Length of second side = 20

Also given the angle between them is 90°.

Hence we can say figure shown is a right angled triangle.

For finding the third side we can use Pythagoras theorem which states that;

"If a triangle is right angled then, the Square of the hypotenuse is equal to sum of the square of the other two sides."

Third \ side ^2=First \ side^2+Second \ side^2

Hence substituting the values we get;

Third\ side^2= 48^2+20^2 = 2304+400 =2704

Now taking Square root on both side we get;

\sqrt{Third \ side^2} = \sqrt{2704}\\ Third \ side = 52 \ units

Hence Length of Third side is 52 units.

3 0
3 years ago
Which of the following expressions is this one equivalent to?
schepotkina [342]

Answer: C. x-2

Step-by-step explanation:

We have the following expression:

\frac{x^{4} -2x^{3} +x-2}{x^{3}+1}

Factoring in the numerator:

\frac{x(x^{3}+1)-2(x^{3}+1)}{x^{3}+1}

Factoring again in the numerator with common factor x^{3}+1:

\frac{(x^{3}+1)(x-2)}{x^{3}+1}

Simplifying:

x-2

Hence, the correct option is C. x-2

3 0
3 years ago
Beth has 30 days to train for the Olympics. She swims 20 laps on the first day, then each day after that she swims two more laps
hjlf
Wait, what’s the question?
7 0
3 years ago
Basic algebra question I struggle with
maxonik [38]
Cross multiply
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