There are 5 dogwood trees currently in the park. Park workers will plant 6

Edna decided to purchase a $15,000 MSRP vehicle at a 5% interest rate for 3 years. The dealership offered her a $1500 cash-back

Ad libitum [116K]

P = A/D, Where P = Monthly payments, A = Total amount owed = 15,000-1,500 = $13,500,

$D= \frac{( 1+ \frac{r}{12} )^{nt} -1}{ \frac{r}{12} (1+ \frac{r}{12}) ^{nt} }$

r = 5% = 0.05, nt = 12*3 = 36

Therefore,
D = $\frac{(1+ 0.05/12)^{36}-1 }{(0.05/12)(1+0.05/12)^{36} }$ = 33.37

Then,

P = 13,500/44.37 = $404.61
The correct answer is c.

5 0

3 years ago

Ben left a 20% tip on his bill at a restaurant. The bill was $40

tigry1 [53]

Answer:

Tip: $8

Total:$48

Step-by-step explanation:

Multiply 40 by 20% to find out how much money he gave for tip.

Add the amount of tip, 8, to $40 find out the total amount of money he spent.

3 0

3 years ago

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The plate, including the bound

rjkz [21]

Answer:

We have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . and the hottest value is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

$T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x$

Let's take the partials derivatives.

$T_{x}(x,y)=2x-\frac{3}{2}=0$

$T_{y}(x,y)=4y=0$

So, we can find the critical point (x,y) of T(x,y).

$2x-\frac{3}{2}=0$

$x=\frac{3}{4}$

$4y=0$

$y=0$

The critical point is (3/4,0) so the temperature at this point is: $T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})$

$T(\frac{3}{4},0)=-\frac{9}{16}$

Now, we need to evaluate the boundary condition.

$x^{2}+y^{2}=1$

We can solve this equation for y and evaluate this value in the temperature.

$y=\pm \sqrt{1-x^{2}}$

$T(x,\sqrt{1-x^{2}})=x^{2}+2(1-x^{2})-\frac{3}{2}x$

$T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2$

Now, let's find the critical point again, as we did above.

$T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0$

$x=-\frac{3}{4}$

Evaluating T(x,y) at this point, we have:

$T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2$

$T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$

Now, we can see that at point (3/4,0) we have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . On the other hand, at the point $-(3/4),\frac{\sqrt{7}}{4})$ we have the hottest value of temperature, it is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

I hope it helps you!

4 0

2 years ago

62 is divisible by which of the following numbers

SVEN [57.7K]

Answer:

D.

Step-by-step explanation:

4 0

3 years ago

In krissy's collection, there are 20 action movies and 28 comedies. what is the ratio of comedies to action movies in krissy's c

Anvisha [2.4K]

28/20. Reduce to 7/5

4 0

3 years ago