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2 years ago

Answer question in image attached :) 20 points

Mathematics

2 answers:

8090 [49]2 years ago

8 0

Answer:

AC = 3(tan 50°)

Step-by-step explanation:

Trig ratios:

$sin(\theta)=\dfrac{O}{H} \ \ \ \ cos(\theta)=\dfrac{A}{H} \ \ \ \ tan(\theta)=\dfrac{O}{A}$

where $\theta$ is the angle, O is the side opposite the angle, A is the side adjacent to the angle and H is the hypotenuse of a right triangle

Given:

Angle = 50°
A = 3 in
O = AC

Therefore, use the tan ratio:

$\implies tan(50)=\dfrac{AC}{3}\\\\\implies AC=3tan(50)\\\\$

REY [17]2 years ago

6 0

Answer:

3(tan 50°)

Step-by-step explanation:

AC is the height

=> tan50 = AC/3

=> AC = 3(tan 50°)

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You purchase a car worth $25077 and make a down payment of $3560. You intend to repay the balance of the car with month end car

zubka84 [21]

Answer:

value of buyout is $4185.74

Step-by-step explanation:

given data

car worth = $25077

down payment = $3560

monthly payment = $336 = 336 × 6 = $2016 per semi annually

time = 5 year = 10 half yearly

rate = 4.04 %

to find out

value of final buyout

solution

we know here loan amount will be 25077 - 3560 = $21517

and we find present value first by formula that is

present value = $\frac{amount(1+r)^t - 1}{r (1+r)^t}$

put here t = 10 and r = $\frac{4.04}{200}$

present value = $\frac{2016(1+\frac{4.04}{200})^10 - 1}{r (1+\frac{4.04}{200})^10}$

present value = 18089.96

loan unpaid amount is here

loan unpaid amount = 21517 - 18089.96

loan unpaid amount = $3427.04

now we calculate value of buyout

that is express as

amount = principal × $(1+r)^{t}$

amount = 3427.04 × $(1+\frac{4.04}{200})^{10}$

amount = 4185.74

so value of buyout is $4185.74

7 0

4 years ago

A tank is filled at a constant rate. 10 minutes after filling is started, the tank contains 4.8L of water. After 35 minutes the

Mashcka [7]

Answer:

Part A)

0.1 liters per minute.

Part B)

There was initially 3.8 liters of water.

$\displaystyle V(t) = 0.1(t - 10) + 4.8$

Part C)

562 minutes.

Step-by-step explanation:

A tank is filled at a constant rate. After 10 minutes, the tank contains 4.8 L of water and after 35 minutes, the tank contains 7.3 L of water.

Part A)

We can represent the current data with two points: (10, 4.8) and (35, 7.3). The x-coordinate is measured in minutes since the tank began to be filled and the y-coordinate is measured in how full the tank is in liters.

To find the rate at which the tank is being filled, find the slope between the two points:

$\displaystyle m = \frac{\Delta y}{\Delta x} = \frac{(7.3)-(4.8)}{(35)-(10)} = \frac{2.5}{25} = 0.1$

In other words, the rate at which the tank is being filled is 0.1 liters per minute.

Part B)

To find the function of the volume of the tank, we can use the point-slope form to first find its equation:

$\displaystyle y - y_1 = m( x - x_1)$

Where m is the slope/rate of change and (x₁, y₁) is a point.

We will substitute 0.1 for m and let (10, 4.8) be the point. Hence:

$\displaystyle y - (4.8) = 0.1(x - 10)$

Simplify:

$\displaystyle y = 0.1(x-10) + 4.8$

Since y represent how full the tank is and x represent the time in minutes since the tank began to be filled, we can substitute y for V(t) and x for t. Thus, our function is:

$\displaystyle V(t) = 0.1(t - 10) + 4.8$

The initial volume is when t = 0. Evaluate:

$\displaystyle V(0) = 0.1 ((0) - 10) + 4.8 = 3.8$

There was initially 3.8 liters of water.

Part C)

To find how long it will take for the tank to be completely filled given its maximum capacity of 60 liters, we can let V(t) = 60 and solve for t. Hence:

$60 = 0.1(t - 10) + 4.8$