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2 years ago

Pls help I’ll brainlest

Mathematics

2 answers:

Flura [38]2 years ago

7 0

3(x+2)-2x=6(x-2)
3x+6-2x=6x-12
1x+6=6x-12
-1x -1x
6=5x-12
+12. +12
18=5x
3.6=x

1.Distribute
2 if you do something to one side of the equation you have to do the same to the other

Marianna [84]2 years ago

6 0

Answer:

Step-by-step explanation:

Mistakes: The student did not distribute the 3 into 2, and the student did not distribute the 6 into negative 2. The student also left out the negative 2x in the first half of the equation.

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I do no understand this

vovangra [49]

The equation of a line for the given slope is $y=\frac{1}{2} x+1$ .

Solution:

Given data:

Slope (m) = $\frac{1}{2}$

Point = (–2, 0)

$x_1=-2, y_1=0$

To find the equation of the given line in the point-slope form:

Point-slope formula:

$y-y_1=m(x-x_1)$

Substitute the given values in the formula.

$$y-0=\frac{1}{2} (x-(-2))$

$$y=\frac{1}{2} (x+2)$

$$y=\frac{1}{2} x+\frac{2}{2}$

$$y=\frac{1}{2} x+1$

The equation of a line for the given slope is $y=\frac{1}{2} x+1$ .

5 0

4 years ago

A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of P lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for S.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for C :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick P such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

5 0

3 years ago

Y=5-1/2x x=-2,0,2,4

telo118 [61]

Answer:

y=6,5,4,3 are value for y

6 0

4 years ago

Double the sum of x and 5

jolli1 [7]

Answer:

2(x+5) = 2x+10

Step-by-step explanation:

2(x+5)

= 2*x + 2*5

= 2x+10

6 0

3 years ago

Consider the absolute value function f(x)=-|x+2|-2 The vertex of the function is

Black_prince [1.1K]

(-2,-2)

Since it has a negative in front of the absolute value, the graph is a downward cone.

3 0

3 years ago

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