A new car is purchased for 17900 dollars. The value of the car depreciates at 13% per

Can someone solve what is n/n^3

emmasim [6.3K]

Answer:

Alex = No B.

Step-by-step explanation:

1) Use Quotient Rule: $\frac{x^{a} }{x^{b} } =x^{a-b}$

$n^{1-3}$

2) Simplify 1 - 3 = 2

$n^{-2}$

3) Use Negative Power Rule: $x^{-a} =\frac{1}{x^{a} }$

$\frac{1}{n^{2} }$

3 0

2 years ago

100 POINTS FOR WHO EVER SHOWS THE WORKS AND ANSWERS CORRECTLY. LEGIT 4 QUETIONS BUT MUST SHOW WORK!!

lianna [129]

Step-by-step explanation:

Hey there!

1.》 Let the unknown angle be 'x'.

Now,

x+ 29° + 105° = 180° [ Sum of interior angle of a triangle is 180°]

x = 180° - 134°

x = 46°

Therefore the missingangle is 46°.

2.》Angle 1 = 78° + 55° [ Exterior angleis equal to the measure of adjacent interior angle]

angle 1 = 133°

Therefore the measure of angle 1 is 133°.

3.》Angle 2 + 55° + 78°= 180° [ Sum of interior angle of a triangle is 180°]

angle 2 = 180° - 133°

angle 2 = 47°

Therefore the measure of angle 2 is 47°.

4.》It's is obvious that the measure of exterior angle is equal to opposite adjacent interior angle of a triangle.

See in picture;;; for more description for no.4.

Hope it helps...

8 0

3 years ago

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a

Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q = -t/100 + c

$Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c} = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}$

when t = 0, Q = 200 L × 1 g/L = 200 g

$Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}$

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

$2 = 200e^{(-t/100)}\\\frac{2}{200} = e^{(-t/100)}$

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0

4 years ago

Pls help! I am really stuck. Find the domain of this graph.

Rudik [331]

Answer:

8

Step-by-step explanation:

The domain is from the points (-4,0) through (4,0).

6 0

3 years ago

Read 2 more answers

The depth of a lake in cedar burg changes over time due to rain fall and evaporation this year it was 39 feet which is 25% less

Nostrana [21]

Answer:

The depth of the lake last year was 52 feet

Step-by-step explanation:

Given as :

This year depth of lake = 39 feet

Let the last year depth of lake = x feet

Now,

According to question :

The depth this year is 25% less than that of last year

I.e 39 feet = x - 25% of x

Or, 39 feet = $\frac{75x}{100}$

or, x = $\frac{3900}{75}$

∴ x = 52 feet

Hence The depth of the lake last year was 52 feet Answer

4 0

3 years ago