Find the value of the following expression:<br> 5 3/5 ÷ 1 1/2 = ????

Which equation can be used to solve for m, the greater integer? m(m – 3) = 108 m(m + 3) = 108 (m + 3)(m – 3) = 108 (m – 12)(m –

nata0808 [166]

Answer:

m(m-3)=108

Step-by-step explanation:

Complete question below:

Two positive integers are 3 units apart on a number line. Their product is 108.

Which equation can be used to solve for m, the greater integer?

m(m – 3) = 108

m(m + 3) = 108

(m + 3)(m – 3) = 108

(m – 12)(m – 9) = 108

Solution

On the number line,

Let

m= larger integer

The integers are 3 numbers apart on the number line, so

m-3=smaller integer

The product (×) of the larger and smaller integers=108

(m)*(m-3)=108

m(m-3)=108

Therefore, the equation that can be used to solve for m, the larger integer is:

m(m – 3) = 108

5 0

3 years ago

A rumor spreads through a small town. Let y(t) be the fraction of the population that has heard the rumor at time t and assume t

sladkih [1.3K]

Answer:

The answer is shown below

Step-by-step explanation:

Let y(t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1−y that has not yet heard the rumor.

a)

$\frac{dy}{dt}\ \alpha\ y(1-y)$

$\frac{dy}{dt}=ky(1-y)$

where k is the constant of proportionality, dy/dt = rate at which the rumor spreads

b)

$\frac{dy}{dt}=ky(1-y)\\\frac{dy}{y(1-y)}=kdt\\\int\limits {\frac{dy}{y(1-y)}} \, =\int\limit {kdt}\\\int\limits {\frac{dy}{y}} +\int\limits {\frac{dy}{1-y}} =\int\limit {kdt}\\\\ln(y)-ln(1-y)=kt+c\\ln(\frac{y}{1-y}) =kt+c\\taking \ exponential \ of\ both \ sides\\\frac{y}{1-y} =e^{kt+c}\\\frac{y}{1-y} =e^{kt}e^c\\let\ A=e^c\\\frac{y}{1-y} =Ae^{kt}\\y=(1-y)Ae^{kt}\\y=\frac{Ae^{kt}}{1+Ae^{kt}} \\at \ t=0,y=10\%\\0.1=\frac{Ae^{k*0}}{1+Ae^{k*0}} \\0.1=\frac{A}{1+A} \\A=\frac{1}{9} \\$

$y=\frac{\frac{1}{9} e^{kt}}{1+\frac{1}{9} e^{kt}}\\y=\frac{1}{1+9e^{-kt}}$

At t = 2, y = 40% = 0.4

c) At y = 75% = 0.75

$y=\frac{1}{1+9e^{-0.8959t}}\\0.75=\frac{1}{1+9e^{-0.8959t}}\\t=3.68\ days$

5 0

2 years ago

Using the data shown on the graph, which statements are correct

Tomtit [17]

Answer:

B, C, and E

Step-by-step explanation:

Let's consider each option given the graph shown above:

A. The ratio of x/y can be found using any point on the line, say, (5, 6). Thus, ratio of x/y = 5/6, not 6/5.

Option A is not correct.

B. If you check for y/x for any given point along the line, we would have the same result. So the ratio of y/x is consistent. Option B is correct.

C. The graph has a line that runs through the point of origin, so therefore, it represents a proportional relationship.

Option C is correct

D. If option C is correct, then option D is not correct.

E. If option C is correct, then option E is correct.

8 0

3 years ago

The function f(x)={1/5}^x is translated up 4 units. Which equation represents the translated function

insens350 [35]

$f(x+n)\to\text{translation n units left}\\\\f(x-n)\to\text{translation n units right}\\\\f(x)+n\to\text{translation n units up}\\\\f(x)-n\to\text{translation n units down}\\\\\\f(x)=\left(\dfrac{1}{5}\right)^x\to f(x)+4=\left(\dfrac{1}{5}\right)^x+4\\\\Answer:\ g(x)=\left(\dfrac{1}{5}\right)^x+4$

3 0

3 years ago

Bacteria X has a growth rate of 180 % per hour. Some amount of bacteria X are accidentally introduced into some egg salad. Three

marta [7]

There were about 6001 bacteria X bring introduced into the egg salad. The number of bacteria in the food 4 hours after contamination is 63000.

6 0

3 years ago

Find the value of the following expression: 5 3/5 ÷ 1 1/2 = ????