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3 years ago

Answers for both boxes please

Mathematics

1 answer:

GuDViN [60]3 years ago

5 0

Answer: (4,2). Explanation: In the elimination method you either add or subtract the equations to get an equation in one variable. When the coefficients of one variable are opposites you add the equations to eliminate a variable and when the coefficients of one variable are equal you subtract the equations to eliminate a variable. The Elimination Method
Step 1: Multiply each equation by a suitable number so that the two equations have the same leading coefficient. ...
Step 2: Subtract the second equation from the first.
Step 3: Solve this new equation for y.
Step 4: Substitute y = 2 into either Equation 1 or Equation 2 above and solve for x.
variable equation. It is relatively difficult to determine the values of x and y without manipulating the equations. If one adds the two equations together, the x s cancel out; the x is eliminated from the problem. Hence it is called the "elimination method”.

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What is the sum of the roots of the quadratic equation x²+6x-14=0?

aivan3 [116]

Answer:

<h3>Sum of Roots =<u> (-8)</u><u> </u><u> </u></h3><h3>Product of Roots = <u> </u><u> </u><u>(-84)</u><u> </u><u> </u></h3>

Step-by-step explanation:

${x}^{2} + 6x - 14 = 0$

Roots :- 6 and -14

Sum of the roots :

[6 + (-14)]

= (-8)//

Product of the roots :

[(6)(-14)]

= (-84)//

8 0

3 years ago

This is confusing<br> (2x+y)^2−(x−2y)^2

Iteru [2.4K]

Answer:

${(2x + y)}^{2} - {(x - 2y)}^{2}$

$< {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} >$

$< {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} >$

${(2x)}^{2} + 2(2x)(y) + {(y)}^{2} - {(x)}^{2} - 2(x)(2y) + {(2y)}^{2}$

${4x}^{2} + 4xy + {y}^{2} - {x}^{2} - 4xy + {4y}^{2}$

${4x}^{2} - {x}^{2} + {4y}^{2} + {y}^{2}$

$= {3x}^{2} + {5y}^{2}$

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3 years ago

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72 = 8 x 9

vivado [14]

8988989889 IDK hate doing the it but luv getting the answers

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3 years ago

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Eva8 [605]

Answer:

e) 2.5

Step-by-step explanation:

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4 years ago

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A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into t

Licemer1 [7]

If $A(t)$ is the amount of salt in the tank at time $t$ , then the rate at which this amount changes over time is given by the ODE

$A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$

$A'(t)+\dfrac1{100}A(t)=15$

We're told that the tank initially starts with no salt in the water, so $A(0)=0$ .

Multiply both sides by an integrating factor, $e^{t/100}$ :

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}$
$\left(e^{t/100}A(t)\right)'=15e^{t/100}$
$e^{t/100}A(t)=1500e^{t/100}+C$
$A(t)=1500+Ce^{-t/100}$

Since $A(0)=0$ , we have

$0=1500+C\implies C=-1500$

so that the amount of salt in the tank over time is given by

$A(t)=1500(1-e^{-t/100})$

After 10 minutes, the amount of salt in the tank is

$A(10)=1500(1-e^{-1/10})\approx142.74\text{ lb}$

8 0

4 years ago

Answers for both boxes please ​

Answers for both boxes please