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3 years ago

2(x-3)(x-3)-8 in standard form

Mathematics

2 answers:

mash [69]3 years ago

7 0

Answer:

2x^2-12x+10

Step-by-step explanation:

(2x-6)(2x-6)-8

2x^2-12x+12-8

2x^2-12x+10

Delvig [45]3 years ago

6 0

Answer:

2x^2-12x+10

Step-by-step explanation:

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Izzy had an 82 and an 88 on his first two tests and believed that he did well enough on his final exam to keep his B average. H

Anastaziya [24]

Answer:

Step-by-step explanation:

to find the mean you need to add up all grades and divide if by the number of terms you added

(82 + 88 + x) ÷ 3 = 69 (since we don't know the third grade let's call it x)

170 + x = 69 × 3

x = 37

if the digits transposed the the actual grade would be 73

5 0

4 years ago

Find the volume of a cylinder with a diameter of 10 inches and a height that is three times the radius. Use 3.14 for pi and roun

maxonik [38]

V = pi r^2 h
v = 3.14 * 5^2 * 15
v = 3.14 * 25 * 15
v = 1177.5 in^3

7 0

3 years ago

Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

$60\cdot \frac{dc}{dt} + 6\cdot c = 3$

$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

$e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C$

$c = 30 + C\cdot e^{-\frac{t}{10} }$

The initial concentration in the tank is:

$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

$C = -29.833$

The solution of the differential equation is:

$c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }$

Now, the quantity of salt at time t is:

$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

Where t is measured in minutes.

7 0

3 years ago

Solve the following equation:<br> 30= |- 4y + 2|

photoshop1234 [79]

Suppose $-4y+2\ge0$ . Then by definition of absolute value,

$|-4y+2| = -4y+2 \implies 30 = -4y+2 \implies -4y = 28 \implies \boxed{y=-7}$

On the other hand, suppose $-4y+2$ . Then

$|-4y+2| = -(-4y+2) \implies 30 = 4y - 2 \implies 4y = 32 \implies \boxed{y=8}$

Recall the definition,

$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$

3 0

2 years ago

This is just probability, please help me !

gtnhenbr [62]

Answer:

4/5

Step-by-step explanation:

Two are white.

10 total marbles.

10 - 2

8 are not white.

8/10

Simplify or reduce.

⇒ 4/5

5 0

3 years ago