All categories

4 years ago

How do I use y=mx to solve this?

Mathematics

1 answer:

stepladder [879]4 years ago

5 0

Answer:

see explanation

Step-by-step explanation:

let m be mars and e be earth.

Given that m varies directly with e then the equation relating them is

m = ke ← k is the constant of variation

To find k use the condition m = 50, e = 150, then

k = $\frac{m}{e}$ = $\frac{50}{150}$ = $\frac{1}{3}$

m = $\frac{1}{3}$ e ← equation of variation

When e = 120, then

m = $\frac{1}{3}$ × 120 = 40 pounds

You might be interested in

Please help!!! ASAP someone please

murzikaleks [220]

Answer:

0.31 yr

Step-by-step explanation:

The formula for interest compounded continuously is

$FV = PVe^{rt}$

FV = future value, and

PV = present value

If FV is twice the PV, we can calculate the doubling time, t

$\begin{array}{rcl}2 & = & e^{rt}\\\ln 2 & = & rt\\t & = & \dfrac{\ln 2}{r} \\\end{array}$

1. Brianna's doubling time

$\begin{array}{rcl}t & = & \dfrac{\ln 2}{0.065}\\\\& = & \textbf{10.663 yr}\\\end{array}$

2. Adam's doubling time

The formula for interest compounded periodically is

$FV = PV\left (1 + \dfrac{r}{n} \right )^{nt}$

where

n = the number of payments per year

If FV is twice the PV, we can calculate the doubling time.

$\begin{array}{rcl}2 & = & \left (1 + \dfrac{0.0675}{4} \right )^{4t}\\\\&= & (1 + 0.016875 )^{4t}\\& = & 1.016875^{4t}\\\ln 2& = & 4 (\ln 1.01688)\times t \\& = & 0.066937t\\t& = & \dfrac{\ln 2}{0.066937}\\\\& = & \textbf{10.355 yr}\\\end{array}$

3. Brianna's doubling time vs Adam's

10.663 - 10.355 = 0.31 yr

It would take 0.31 yr longer for Brianna's money to double than Adam's.

3 0

3 years ago

Mr. Singh just left his house to go to work and has a meeting as soon as he arrives. The

vodomira [7]

Answer:5

Step-by-step explanation:

3 0

4 years ago

One method of slowing the growth of an insect population without using pesticides is to introduce into the population a number o

Zielflug [23.3K]

To be clear, the given relation between time and female population is an integral:
 $t = \int { \frac{P+S}{P[(r - 1)P - S]} } \,
dP$

The problem says that r = 1.2 and S = 400, therefore substituting:
 $t = \int { \frac{P+400}{P[(1.2 - 1)P - 400]}
} \, dP$ 

= $
\int { \frac{P+400}{P(0.2P - 400)} } \, dP$

In order to evaluate this integral, we need to write this rational function in a simpler way:
 $\frac{P+400}{P(0.2P - 400)} = \frac{A}{P} +
\frac{B}{(0.2P - 400)}$ 

where we need to evaluate A and B. In order to do so, let's calculate the LCD:
 $\frac{P+400}{P(0.2P - 400)} = \frac{A(0.2P -
400)}{P(0.2P - 400)} + \frac{BP}{P(0.2P - 400)}$ 

the denominators cancel out and we get:
P + 400 = 0.2AP - 400A + BP
 = P(0.2A + B) - 400A

The two sides must be equal to each other, bringing the system:
 $\left \{ {{0.2A + B = 1} \atop {-400A =
400}} \right.$ 

Which can be easily solved:
 $\left \{ {{B=1.2} \atop {A=-1}} \right.$ 

Therefore, our integral can be written as:
 $t = \int { \frac{P+400}{P(0.2P - 400)} } \,
dP = \int {( \frac{-1}{P} + \frac{1.2}{0.2P-400} )} \, dP$ 
= $- \int { \frac{1}{P} \, dP +
1.2\int { \frac{1}{0.2P-400} } \, dP$ 
= $- \int { \frac{1}{P} \, dP +
6\int { \frac{0.2}{0.2P-400} } \, dP$ 
= - ln |P| + 6 ln |0.2P - 400| + C

Now, let’s evaluate C by considering that at t = 0 P = 10000:
0 = - ln |10000| + 6 ln |0.2(10000) - 400| + C
C = ln |10000| - 6 ln |1600|
C = ln (10⁴) - 6 ln (2⁶·5²)
C = 4 ln (10) - 36 ln (2) - 12 ln (5) 
 Therefore, the equation relating female population with time requested is:
t = - ln |P| + 6 ln |0.2P - 400| + 4 ln (10) - 36 ln (2) - 12 ln (5)

8 0

3 years ago

In the expression 8k + 5. which best describes what 8 represents?

ZanzabumX [31]

Answer:

it represents a variable

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

There are 2.54 centimeters in an 1 inch. how many centimeters are 17 inches

Tpy6a [65]