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3 years ago

How many 0.2-liter glasses of water are contained in a 3.6-liter pitcher

Mathematics

2 answers:

jolli1 [7]3 years ago

4 0

18 glasses of water

3.6/.2=18

Alexandra [31]3 years ago

3 0

<h2>Answer:</h2>

<u>The answer is </u><u>18 glasses</u>

<h2>Step-by-step explanation:</h2>

Total amount in liters in pitcher = 3.6

Amount of one glass in liters = 0.2 liters

Amount of glasses in pitcher will be

3.6 / 0.2

Which comes out to be 18

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If n=4 the value of the expression 28-12+ square root of n divided by 2 is____.

Ann [662]

C 17/2 hope this helps you

4 0

3 years ago

What is the domain of f(x) = (x + 2)2 – 1?

larisa [96]

Simplify

$\\ \tt\Rrightarrow f(x)=(x+2)^2-1$

$\\ \tt\Rrightarrow f(x)=x^2+4x+2-1$

$\\ \tt\Rrightarrow f(x)=x^2+4x+1$

The Domain is set of all real numbers

$\\ \tt\Rrightarrow Domain\in R$

4 0

2 years ago

You have a wire that is 20 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The o

Aleksandr [31]

Answer:

Therefore the circumference of the circle is $=\frac{20\pi}{4+\pi}$

Step-by-step explanation:

Let the side of the square be s

and the radius of the circle be r

The perimeter of the square is = 4s

The circumference of the circle is =2πr

Given that the length of the wire is 20 cm.

According to the problem,

4s + 2πr =20

⇒2s+πr =10

$\Rightarrow s=\frac{10-\pi r}{2}$

The area of the circle is = πr²

The area of the square is = s²

A represent the total area of the square and circle.

A=πr²+s²

Putting the value of s

$A=\pi r^2+ (\frac{10-\pi r}{2})^2$

$\Rightarrow A= \pi r^2+(\frac{10}{2})^2-2.\frac{10}{2}.\frac{\pi r}{2}+ (\frac{\pi r}{2})^2$

$\Rightarrow A=\pi r^2 +25-5 \pi r +\frac{\pi^2r^2}{4}$

$\Rightarrow A=\pi r^2\frac{4+\pi}{4} -5\pi r +25$

For maximum or minimum $\frac{dA}{dr}=0$

Differentiating with respect to r

$\frac{dA}{dr}= \frac{2\pi r(4+\pi)}{4} -5\pi$

Again differentiating with respect to r

$\frac{d^2A}{dr^2}=\frac{2\pi (4+\pi)}{4}$ > 0

For maximum or minimum

$\frac{dA}{dr}=0$

$\Rightarrow \frac{2\pi r(4+\pi)}{4} -5\pi=0$

$\Rightarrow r = \frac{10\pi }{\pi(4+\pi)}$

$\Rightarrow r=\frac{10}{4+\pi}$

$\frac{d^2A}{dr^2}|_{ r=\frac{10}{4+\pi}}=\frac{2\pi (4+\pi)}{4}>0$

Therefore at $r=\frac{10}{4+\pi}$ , A is minimum.

Therefore the circumference of the circle is

$=2 \pi \frac{10}{4+\pi}$

$=\frac{20\pi}{4+\pi}$

4 0

2 years ago

In exercises 8-10, write a system of linear equations that has the ordered pair as its solution. (-6, -2)

Anika [276]

Answer:

(8) $x + y = -8$ and $2x - y = -10$

(9) $3x - y = -54$ and $4x + y = -30$

(10) $x + y = 2$ and $x - y = 2$

Step-by-step explanation:

Given [Missing from the question]

$(8).\ (-6,-2)\ \\ (9).\ (-12,18)\ \\ (10).\ (2,0)$

Required

Write a system of linear equations

The solutions to this question is open and have different solutions

Solving (8) (-6,-2)

Let the equations be: x + y and 2x - y

$x + y = -6 - 2$

$x + y = -8$ --- (1)

$2x - y = 2 * (-6) - (-2)$

$2x - y = -10$ -- (2)

So, the equations are:

$x + y = -8$ and $2x - y = -10$

Solving (9) (-12, 18)

Let the equations be: 3x - y and 4x + y

$3x - y = 3 *(-12) - 18$

$3x - y = -54$

$4x + y = 4 * (-12) + 18$

$4x + y = -30$

So, the equations are:

$3x - y = -54$ and $4x + y = -30$

Solving (9): (2,0)

Let the equations be: x + y and x - y

$x + y = 2 + 0$

$x + y = 2$

$x - y = 2 - 0$

$x - y = 2$

So, the equations are:

$x + y = 2$ and $x - y = 2$

5 0

2 years ago

A summer music program admits a total of 30 students. Every year the program is filled. Some years there are only girls, and som

vesna_86 [32]

Answer:

c d

Step-by-step explanation:

trust me

5 0

3 years ago