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3 years ago

Would the answer be .0115 Km?

Mathematics

2 answers:

uysha [10]3 years ago

7 0

Yes it would be 0.115 km.

RUDIKE [14]3 years ago

5 0

We know, 1 km = 1000 m
so, 1 m = 1/1000 Km

Now, 115 m = 1/1000 * 115 = 0.115 Km

In short, Your Answer would be: 0.115 Km

Hope this helps!

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Weird math question.

Aneli [31]

Answer:

It is 4/11

Step-by-step explanation:

Let x = 0.363636363636

Now as there are two digits repeating immediately after decimal point, we multiply above by 100 and get

100x = 36.363636363636

Now subtracting first equation from second we get

99x = 36

and hence

x = 36/ 99

= (9 × 4) / (9 × 11)

which then equals

= 4/11

6 0

3 years ago

-4(d+5)-3d>8 please help i dont need you to explain it only answer please and thank you

Gala2k [10]

Show your work kidddd

8 0

3 years ago

Part : Find g(x) by identifying the function you are taking the square root of

aksik [14]

Answer:

g(x) = (3x-4)

Step-by-step explanation:

F(x) = sqrt( 3x-4)

We are taking the square root of (3x-4)

so g(x) = (3x-4)

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

#31 please help and show work thank you in advance

antiseptic1488 [7]

The base price of the car is about $22,196 because:

x= major options package

10x + x = 24,416

*the 10x is the price of the car

This then simplifies to

11x = 24,416

use division.

The destination charge is not needed unless you want to check your answer. If you have to then you would do the same thing. The equation would be:

y= destination charge

5y + y = 24,416

Hope this helps.

6 0

3 years ago