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oee [108]
2 years ago
7

The height of

Mathematics
1 answer:
Amiraneli [1.4K]2 years ago
4 0

Hey there! I'm happy to help!

This glass is most likely a cylinder. To find the volume of a cylinder you take the area of the circle base and multiply it by the height.

To find the area of a circle, you square the radius and multiply it by pi (we will use 3.14 for pi).

Our radius is 3cm.

3

We square it.

3²=9

Multiply by 3.14.

9×3.14=28.26

Now that we have our circle base area, we multiply it by the height (7 cm).

28.26×7= 197.82

This is how many cubic centimeters the glass can hold if it is completely full. However, we want half full, so we can just divide by 2.

197.82/2=98.91

We can also multiply our circle base by 3.5, which will give us half of the full height.

So, the height of the water is 3.5 cm and the volume of the water is 98.1 cm³.

Have a wonderful day and keep on learning! :D

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Even number less than 12​
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Answer:

2,4,6,8,10

Step-by-step explanation:

they are even numbers

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3 years ago
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Bryant brenda and jack ate lunch together. Bryant ate 1/3 of a pizza . brenda ate 1/2 of what bryant ate Jack ate 1/2 a pizza mo
sergejj [24]

Answer:

2/3 for Jack

Step-by-step explanation: You know Bryant eat 1/3 of a pizza, so you got 2/3 for Jack because Brenda is 1/3 since thats half the amount Jack had. :)

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2 years ago
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What number can be added to -2 to get a sum of 0?
levacccp [35]







Answer = 2
|-------------------|-----------------|
5 0
3 years ago
HELP ASAP PLZ i need help !<br>help me :((
kodGreya [7K]

Answer:

(1) g[f(x)]=\frac{8x-1}{12x-4}

(2)  g^{-1}(x)=\frac{x+1}{(3x-2)}

(3)    x=\frac{9+\sqrt{129} }{6},  \frac{9-\sqrt{129} }{6}

Step-by-step explanation:

Given functions f(x) = (4x-1)

are                     g(x)=\frac{2x+1}{3x-1}

(1)  g[f(x)]=\frac{2(4x-1)+1}{3(4x-1)-1}

                =\frac{8x-2+1}{12x-3-1}

                =\frac{8x-1}{12x-4}

(2) for g^{-1}(x), rewrite the function g(x) in terms of an equation

y=\frac{2x+1}{3x-1}

Substitute y in place of x and x in place of y, then solve for y.

x=\frac{2y+1}{3y-1}

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

3xy - x = 2y + 1

3xy - 2y = x + 1

y(3x-2) = x + 1

y=(\frac{x+1}{3x-2} )

⇒ g^{-1}(x)=\frac{x+1}{(3x-2)}

(3)    f[g(x)]=(x-1)

       f[g(x)]=4[\frac{2x+1}{3x-1}] =(x-1)

⇒    \frac{(8x+4)-(3x-1)}{3x-1}=(x-1)

⇒    \frac{8x-3x+4+1}{3x-1}=(x-1)

⇒    \frac{5x+5}{(3x-1)}=(x-1)

⇒    5x+5=(3x-1)(x-1)

⇒    5x+5=3x^2-3x-x+1

⇒   5x+5=3x^2-4x+1

⇒   3x^2-9x-4=0

⇒    x=\frac{9\pm\sqrt{(-9)^2-4(-4)\times3} }{2\times3}

   x=\frac{9\pm\sqrt{81+48} }{6}

   x=\frac{9\pm\sqrt{129} }{6}

   x=\frac{9+\sqrt{129} }{6},  \frac{9-\sqrt{129} }{6}

4 0
3 years ago
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By the confront theorem we know that the limit only exists if both lateral limits are equal

In this case they aren't so we don't have limit for x approaching 2, but we can find their laterals.

Approaching 2 by the left we have it on the 5 line so this limit is 5

Approaching 2 by the right we have it on the -3 line so this limit is -3

Think: it's approaching x = 2 BUT IT'S NOT 2, and we only have a different value for x = 2 which is 1, but when it's approach by the left we have the values in the 5 line and by the right in the -3 line.

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3 years ago
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