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3 years ago

Y=−2x+4y, Complete the missing value in the solution to the equation. ( , -2)

Mathematics

1 answer:

viva [34]3 years ago

7 0

-(3/2) is the best answer

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Which system is equivalent to

levacccp [35]

Answer:

Step-by-step explanation:

4 0

3 years ago

If the product of the integers a, b, and c is 1, then what is the difference between the largest and the smallest possible

stepladder [879]

Since there are no multiples other than 1 and -1 that could multiply in any way to equal 1, those are the only numbers you can really work with

it is a possibility that a, b, and c could equal 1 but then it would go wrong when comes time to subtract since all numbers are equal and cannot subtract largest from smallest.

a logical approach would be to try -1, -1, and 1 since -1 * -1 = 1 and then 1* 1 =1

we can make a= -1 , b= -1, and c= 1
(we need b to equal -1 so that we get an answer of -1 opposed to if we made a=-1 and c=-1, all outcomes would equal 1)
-1²=1
-1³=-1
1⁴= 1
since 1 is the largest possible outcome and -1 is the lowest possible outcome you should subtract the two

1-(-1) = 1+1 = 2

The answer should be D: 2

hoped this helped at all!

5 0

3 years ago

Write a linear equation in standard form that satisfies the given set of conditions. X-intercept: 3, y-intercept: 5

Artyom0805 [142]

The linear equation in standard form is $y=\frac{-5}{3}x +5$ .

<h3>Linear Function</h3>

An equation can be represented by a linear function. The standard form for the linear equation is: ax+b , for example, y=7x+2. Where:

a= the slope. It can be calculated for $\frac{\Delta x }{\Delta y}$ .

b= the constant term that represents the y-intercept.

The question gives: X-intercept:3 and y-intercept: 5. Then,

The x-intercept is the point that y=0, then the x-intercept point is (3,0).
The y-intercept is the point that x=0, then the x-intercept point is (0,5).

With this information, you can find the slope (a).

$a=\frac{\Delta y }{\Delta x}= \frac{0-5}{3-0} =\frac{-5}{3}$

The question gives the coefficient b since it gives the y-intercept=5.

Therefore the linear equation is : $y=\frac{-5}{3}x +5$ .

Read more about the linear equations here:

brainly.com/question/2030026

#SPJ1

4 0

2 years ago

5a - 9 + a = 9

Ne4ueva [31]

Answer:

a = 3

Step-by-step explanation:

1) Simplify 5a - 9 + a to 6a - 9.

$6a - 9 = 9$

2) Add 9 to both sides.

$6a = 9 + 9$

3) Simplify 9 + 9 to 18.

$6a = 18$

4) Divide both sides by 6.

$a = \frac{18}{6}$

5) Simplify 18/6 to 3.

$a = 3$

Therefor, the answer is, a = 3.

7 0

3 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

2 years ago