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3 years ago

What is the equation simplified

Mathematics

1 answer:

Lilit [14]3 years ago

4 0

$\bf \cfrac{2}{7}(2s+3)\implies \stackrel{\textit{distributing}}{\cfrac{2}{7}\cdot 2s+\cfrac{2}{7}\cdot 3}\implies \cfrac{2\cdot 2}{7}s+\cfrac{2\cdot 3}{7}\implies \cfrac{4}{7}s+\cfrac{6}{7}$

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What is the center and radius of the circle that has the equation (x + 7)^ 2 + (y − 4) ^2 = 49? Explain.

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Step-by-step explanation:

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Radius: 7

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3 years ago

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Theequationforlinegcanbewrittenas y = –9 4 x − 1 . Line h , whichisperpendiculartoline g , includesthepoint (10, 4) . What isthe

KATRIN_1 [288]

Answer:

y-4 = 4/9(x-10)

Step-by-step explanation:

Given the equation of the line g to be y = -9/4 x - 1.

To find the equation of a line h perpendicular to this line, the following steps must be taken;

First find the slope of the known line g

The standard equation of a line is y = mx+c where

m is the slope

c is the intercept

Comparing this equation with the given equation y = -9/4 x -1, it can be seen that the slope of the line g is -9/4

Next is to find the slope of the line h

Since the line g is perpendicular to h, then the product of their slope will be -1 i.e mh*mg = -1

mh = -1.mg

mh = -1/(-9/4)

mh = -1*-4/9

mh = 4/9

Hence the slope of the line h is 4/9

Find the intercept of line h

To get this, you will substitute the slope m and the point into the equation y = mx+c

m = 4/9

(x,y) = (10,4)

4 = 4/9(10) + c

4 = 40/9 + c

c = 4-40/9

c = (36-40)/9

c = -4/9

Hence the intercept of the line h is -4/9

Finally, find the equation of the line h in slope-intercept form

The slope-intercept form of a line is expressed as y = mx+c

y-y0 = m(x-x0)

Given

m = -4/9

x0 = 10

y0 = 4

Substitute:

y-4 = 4/9(x-10)

Hence the equation in slope-intercept form is y-4 = 4/9(x-10)

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3 years ago

The spinners shown below are each spun one time. A spinner with 3 equal sections labeled 1 through 3.A spinner with 4 equal sect

jenyasd209 [6]

Answer:

the probability of both spinners land on 2 is 1/12

Step-by-step explanation:

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3 years ago

What is the product? startfraction 4 k 2 over k squared minus 4 endfraction times startfraction k minus 2 over 2 k 1 endfraction

Tcecarenko [31]

The product of the expression $(\frac{4k+2}{k^2-4})* \frac{k-2}{2k+1}$ is the expression $\frac{2}{k+2}$

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Given the expression:

$(\frac{4k+2}{k^2-4})* \frac{k-2}{2k+1} \\\\This\ can\ be\ simplified\ to:\\\\(\frac{2(2k+1)(k-2)}{(k+2)(k-2)(2k+1)}) \\\\=\frac{2}{k+2}$

The product of the expression $(\frac{4k+2}{k^2-4})* \frac{k-2}{2k+1}$ is the expression $\frac{2}{k+2}$

Find out more on equation at: brainly.com/question/2972832

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2 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

aliya0001 [1]

Answer:

$A=1500-1450e^{-\dfrac{t}{250}}$

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

$R_{in}$ =(concentration of salt in inflow)(input rate of brine)

$=(2\frac{lbs}{gal})( 3\frac{gal}{min})\\R_{in}=6\frac{lbs}{min}$

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

Concentration, $C(t)=\dfrac{Amount}{Volume}=\dfrac{A(t)}{500}$

$R_{out}$ =(concentration of salt in outflow)(output rate of brine)

$=(\frac{A(t)}{500})( 2\frac{gal}{min})\\R_{out}=\dfrac{A}{250}$

Now, the rate of change of the amount of salt in the tank

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{250}$

We solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{250}=6\\$The integrating factor: e^{\int \frac{1}{250}dt} =e^{\frac{t}{250}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{250}}+\dfrac{A}{250}e^{\frac{t}{250}}=6e^{\frac{t}{250}}\\(Ae^{\frac{t}{250}})'=6e^{\frac{t}{250}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}$

Recall that when t=0, A(t)=50 (our initial condition)

$50=1500+Ce^{-\frac{0}{250}}50=1500+Ce^{0}\\C=-1450\\$Therefore the amount of salt in the tank at any time t is:\\A=1500-1450e^{-\dfrac{t}{250}}$

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3 years ago