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Varvara68 [4.7K]
3 years ago
10

Solve the equation for all real solutions in simplest form. z^2 - 12z +9= -3

Mathematics
1 answer:
Mars2501 [29]3 years ago
3 0

Answer:

Solving the equation for all real solutions in simplest form.

z^2 - 12z +9= -3 we get \mathbf{z=10.9\: or\: z=1.1}

Step-by-step explanation:

We need to solve the equation for all real solutions in simplest form.

z^2 - 12z +9= -3

First simplifying the equation:

z^2 - 12z +9+3= -3+3\\z^2 - 12z +12= 0

Now, we can solve the equation using quadratic formula:

z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

we have a = 1, b=-12 and c=12

Putting values in formula and finding values of x

z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\z=\frac{-(-12)\pm\sqrt{(-12)^2-4(1)(12)}}{2(1)}\\z=\frac{12\pm\sqrt{144-48}}{2}\\z=\frac{12\pm\sqrt{96}}{2}\\z=\frac{12\pm9.8}{2}\\z=\frac{12+9.8}{2}\:or\:z=\frac{12-9.8}{2}\\z=10.9\:or\:z=1.1

So, we get value of z: z=10.9 or z=1.1

Solving the equation for all real solutions in simplest form.

z^2 - 12z +9= -3 we get \mathbf{z=10.9\: or\: z=1.1}

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Answer:

dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0

Step-by-step explanation:

The mass flow rate dQ(t)/dt = mass flowing in - mass flowing out

Since 5 g/L of salt is pumped in at a rate of 4 L/min, the mass flow in is thus 5 g/L × 4 L/min = 20 g/min.

Let Q(t) be the mass present at any time, t. The concentration at any time ,t is thus Q(t)/volume = Q(t)/10. Since water drains at a rate of 4 L/min, the mass flow out is thus, Q(t)/10 g/L × 4 L/min = 2Q(t)/5 g/min.

So, dQ(t)/dt = mass flowing in - mass flowing out

dQ(t)/dt = 20 g/min - 2Q(t)/5 g/min

Since the salt just begins to be pumped in, the initial mass of salt in the tank is zero. So Q(0) = 0

So, the initial value problem is thus

dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0

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Ari buying 10 1/2 feet of rope that costs $5.25 per foot.How much will Ari spend?
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We can use linear combinations of the equations to eliminate variables.

3x - 4y = 1
-2x + 3y = 1

To eliminate y we'll make the linear combination of 3 times the first equation minus four times the second. 

9x - 12y = 3
-8x + 12y = 4

Adding,
x = 7

We could solve for y directly but let's use another linear combination, twice the first plus three times the second:

2(3x - 4y) + 3(-2x + 3y)= 2(1)+3(1)

y = 5

Check: 3(7)-4(5)=1 good.   -2(7)+3(5)=1 good.

Q18 Answer: (7,5)

y = -3x + 5
5x - 4y = -3

4y +1(5x - 4y) = 4(-3x + 5) + 1(-3)

5x = -12x + 20 -3

17 x = 17

x = 1
y = -3(1) + 5 = 2
Check: 5(1) - 4(2) = -3 good

Q19 Answer (1,2)

6x + 5y = 25
x = 2y + 24
6x = 12y + 144
5y = 25 - 12y - 144
17y = -119
y = -119/17= -7
x = 2y+24= 10
Check: 6(10)+5(-7)=25 good   2y+24=2(-7)+24=10=x good

Q20 Answer (10,-7)


3x + y = 18
-7x + 3y = -10
9x  + 3y = 54
9x - -7x = 54 - -10
16x = 64
x=4
y = 18 -3x = 18-12=6

Check: 3(4)+6=18 good,  -7(4)+3(6)=-10 good

Q21 Answer: (4,6)

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