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3 years ago

IaI − IbI + IcI , if a=−8; b=−5; c=1pLEEEEEEZE hELP

Mathematics

1 answer:

yawa3891 [41]3 years ago

6 0

Answer:

Step-by-step explanation:

to solve | a | - | b | + | c | we need to plug in the given values

a = -8

b = -5

c = 1

the | | bars mean absolute value which is the exact value a number is away from 0 on a number line. an absolute value has to be positive, not a negative, as you cant be negative spaces away from something. plugging in the values into the equation we have:

| -8 | - | -5 | + | 1 |

the absolute value of -8 is 8, as -8 is 8 units away from 0 on a number line

like wise with -5, the absolute value of -5 is 5, as -5 is 5 units away from 0 on a number line

1 is the absolute value of 1 as its 1 unit away from 0 on a number line

the new expression looks like the following:

8 - 5 + 1

now we just add/subtract

8 -5 = 3

3 + 1 = 4

the solution to this expression is 4

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LenaWriter [7]

Answer:

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3 years ago

Jake is solving x^2-6x+5=0 by completing the square his steps are shown below step 1. x^2-6x=-5 step 2. x^-6x+9=-5+9 step 3. (x-

damaskus [11]

Answer:

Jake's error in step 3

Step-by-step explanation:

we have

$x^{2} -6x+5=0$

Complete the square

step 1

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2} -6x=-5$

step 2

Complete the square. Remember to balance the equation by adding the same constants to each side

$x^{2} -6x+9=-5+9$

$x^{2} -6x+9=4$

step 3

Rewrite as perfect squares

$(x-3)^{2}=4$

Jake's error in step 3

He placed 6 instead of 3 in the left side

step 4

take square root both sides

$\sqrt{(x-3)^2} =\sqrt{4}$

step 5

$(x-3)=\pm2$

step 6

$x=3\pm2$

step 7

x=5 and x=1

7 0

3 years ago

A tank with a capacity of 1000 L is full of a mixture of water and chlorine with a concentration of 0.02 g of chlorine per liter

faltersainse [42]

At the start, the tank contains

(0.02 g/L) * (1000 L) = 20 g

of chlorine. Let c (t ) denote the amount of chlorine (in grams) in the tank at time t .

Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

(c (t )/(1000 + (10 - 25)t ) g/L) * (25 L/s) = 5c (t ) /(200 - 3t ) g/s

In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after t s there will be (1000 + 10t ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time t is (1000 - 15t ) L. Then the concentration of chlorine per unit volume is c (t ) divided by this volume.

So the amount of chlorine in the tank changes according to

$\dfrac{\mathrm dc(t)}{\mathrm dt}=-\dfrac{5c(t)}{200-3t}$

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5t )^(5/3), then integrate both sides to solve for c (t ):

$\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0$