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3 years ago

Find the area of the following figure if a line drawn from the side of length 8 to the side of length 2 is 8 units long.

Mathematics

1 answer:

Dima020 [189]3 years ago

4 0

Multiply the length and with 4 4 5&5 and add them all up to get the answer

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I dont have a question for this. look at the photo

Artemon [7]

Answer:

first is no second is yes third is no fourth is yes

Step-by-step explanation:

I'm not sure btw

7 0

2 years ago

There are 26 gold fish in the tank at the store. how many ways can ben choose 5?

Lapatulllka [165]

Answer:

Step-by-step explanation:

26/5= 5.2

the greatest whole number that can be chosen is 5.

6 0

3 years ago

Find the circumference of a circle with radius, r = 6.8m. Give your answer rounded to 1 DP.

Assoli18 [71]

Answer:

42.7 m

Step-by-step explanation:

The Circumference of a circle is found by multiplying the radius of the circle by 2π. Here the radius is 6.8 m. The circumference of this circle is

C = 2π(6.8 m) = 13.6π m, or (to 1 decimal place, 42.7 m.

4 0

3 years ago

A certain college graduate borrows 7864 dollars to buy a car. The lender charges interest at an annual rate of 13%. Assuming tha

White raven [17]

Answer:

Therefore rate of payment = $ 3145.72

Therefore the rate of interest = =$1573.17

Step-by-step explanation:

Consider A represent the balance at time t.

A(0)=$ 7864.

r=13 % =0.13

Rate payment = $k

The balance rate increases by interest (product of interest rate and current balance) and payment rate.

$\frac{dB}{dt} = rB-k$

$\Rightarrow \frac{dB}{dt} - rB=-k$ .......(1)

To solve the equation ,we have to find out the integrating factor.

Here p(t)= the coefficient of B =-r

The integrating factor $=e^{\int p(t) dt$

$=e^{\int (-r)dt$

$=e^{-rt}$

Multiplying the integrating factor the both sides of equation (1)

$e^{-rt}\frac{dB}{dt} -e^{-rt}rB=-ke^{-rt}$

$\Rightarrow e^{-rt}dB - e^{-rt}rBdt=-ke^{-rt}dt$

Integrating both sides

$\Rightarrow \int e^{-rt}dB -\int e^{-rt}rBdt=\int-ke^{-rt}dt$

$\Rightarrow e^{-rt}B=\frac{-ke^{-rt}}{-r} +C$ [ where C arbitrary constant]

$\Rightarrow B(t)=\frac{k}{r} +Ce^{rt}$

Initial condition B=7864 when t =0

$\therefore 7864= \frac{k}{r} - Ce^0$

$\Rightarrow C= \frac{k}{r} -7864$

Then the general solution is

$B(t)=\frac{k}{r}-( \frac{k}{r}-7864)e^{rt}$

To determine the payment rate, we have to put the value of B(3), r and t in the general solution.

Here B(3)=0, r=0.13 and t=3

$B(3)=0=\frac{k}{0.13}-( \frac{k}{0.13}-7864)e^{0.13\times 3}$

$\Rightarrow- 0.48\frac{k}{0.13} +11614.98=0$

⇒k≈3145.72

Therefore rate of payment = $ 3145.72

Therefore the rate of interest = ${(3145.72×3)-7864}

=$1573.17

4 0

3 years ago

For what value of x is the value of f(x)=4x-2 equal to 18

mina [271]

F(x) = 4(5) - 2 → f(x) = 20 - 2 → f(x) = 18

5 0

3 years ago

Read 2 more answers