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Pepsi [2]
3 years ago
13

Dog water ur bad kid get better

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
7 0

don't understand your question

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Rewrite the equation by completing the square.<br>x^2-2x+1=0​
kaheart [24]

Answer:

(x - 1)² = 0

Step-by-step explanation:

Given

x² - 2x + 1 = 0 ( subtract 1 from both sides )

x² - 2x = - 1

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(- 1)x + 1 = - 1 + 1

(x - 1)² = 0

5 0
3 years ago
Binks needs to raise at least $3600. He was given $200 and plans to save $50 a week. Select all of the inequalities that represe
kherson [118]

Answer:

I'm pretty sure it is 50n+200 <u><</u> 3600

Step-by-step explanation:

Im to fully sure tho

5 0
3 years ago
1
Sloan [31]

Answer:

<h3>#1</h3>

<u>Rectangle area:</u>

  • A = lw

<u>Given:</u>

  • l = 12 ft, A = 300 ft²

<u>Find w:</u>

  • 12w = 300
  • w = 300/12
  • w = 25 ft
<h3>#2</h3>

<u>Perimeter is the sum of side lengths:</u>

P = 2 1/4 + 5 2/5 + 5 17/20 =

     2 + 5 + 5 + 1/4 + 2/5 + 17/20 =

     12 + 5/20 + 8/20 + 17/20 =

     12 + 30/20 =

     12 + 1.5 =

     13.5 ft

<h3>#3</h3>
  • Cost of a shirt = $19.99
  • Number of shirts = 6

<u>Total cost:</u>

  • $19.99*6 = $119.94

8 0
3 years ago
What can u use to float a boat
Mamont248 [21]
Air? Unless you mean like a toy boat? IDK, I do know that huge ships like the Titanic are able to float because they have trapped oxygen. Think about if you're swimming; often when you hold your breath you float a bit better than if you were heavily breathing. Sorry i know its not alot to go off of, did i help you a little?
5 0
3 years ago
Read 2 more answers
The volume of a rectangular box with a square base remains constant at 500 cm3 as the area of the base increases at a rate of 6
Andre45 [30]

Answer:

the rate at which the height of the box is decreasing is -0.0593 cm/s

Step-by-step explanation:

Given the data in the question;

Constant Volume of a rectangular box with a square base = 500 cm³

area of the base increases at a rate of 6 cm²/sec

so change in the area of the base with respect to time dA/dt = 6 cm²/sec

each side of the base is 15 cm long

so Area of the base = 15 cm × 15 cm = 225 cm²

the rate at which the height of the box is decreasing = ?

Now,

V = Ah

dv/dt = 0 ⇒ Adh/dt + hdA/DT = 0

⇒ dh/dt = -hdA/dt / A

we substitute

dh/dt = [ -( 500 / 225 ) × 6 ] / 225

dh/dt = [ -(2.22222 × 6)  ] / 225

dh/dt = [ -13.3333 ] / 225

dh/dt = -0.0593 cm/s

Therefore, the rate at which the height of the box is decreasing is -0.0593 cm/s

7 0
3 years ago
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