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horsena [70]
2 years ago
6

A​ cone-shaped paper water cup has a height of 12 cm and a radius of 6 cm. If the cup is filled with water to one-sixth its​ hei

ght, what portion of the volume of the cup is filled with​ water?
Mathematics
1 answer:
Liono4ka [1.6K]2 years ago
8 0

Answer:

  1/216

Step-by-step explanation:

The ratio of volumes of similar shapes is the cube of the scale factor. The filled portion of the cone is similar to the entire cone.

<h3>Linear scale factor</h3>

If the filled height is 1/6 of the total height, the scale factor for linear dimensions is 1/6.

<h3>Volume scale factor</h3>

The scale factor for the volume is the cube of the scale factor for linear dimensions. it is ...

  (1/6)³ = 1/216

The volume of the filled portion of the cup is 1/216 of the volume of the cup.

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Help?? With working out please
andriy [413]

Answer:

Step-by-step explanation:

move all the terms to the right so your equation will equal 0. then look for the common denominator. which in this case is 60 (5*4*3).

multiply all terms by 60 then simplify.

12(2x-3)-15(6-3x)+120-20(7x)=0

24x-36-90+45x+120-140x=0

combine like terms and then divide

-71x=6

x=-6/71

(do not forget parenthesis when simplifying they are very important when it comes to distributing the minus sign)

   

8 0
3 years ago
If x+y=12 and xy=15,find the value of (x^2+y^2)
Mice21 [21]

Answer:  x^2+y^2=\dfrac{105\pm 18\sqrt6}{2}

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

EQ1:  x + y = 12     --> x = 12 - y

EQ2:  xy = 15      

Substitute x = 12-y into EQ2 to solve for y:

(12 - y)y = 15

12y - y² = 15

0 = y² - 12y + 15

    ↓     ↓         ↓

  a=1   b= -12   c=15

.\  y=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\.\quad =\dfrac{-(-12)\pm \sqrt{(-12)^2-4(1)(15)}}{2(1)}\\\\\\.\quad =\dfrac{12\pm \sqrt{144-120}}{2}\\\\\\.\quad =\dfrac{12\pm \sqrt{24}}{2}\\\\\\.\quad =\dfrac{12\pm 2\sqrt{6}}{2}\\\\\\.\quad =6\pm \sqrt{6}

Now, let's solve for x:

xy=15\\\\x(6\pm\sqrt6)=15\\\\x=\dfrac{15}{6\pm\sqrt6}\\\\\\x=\dfrac{15}{6\pm\sqrt6}\bigg(\dfrac{6\pm\sqrt6}{6\pm\sqrt6}\bigg)=\dfrac{6\pm \sqrt6}{2}

Lastly, find x² + y² :

y^2=(6\pm \sqrt6)^2\quad \rightarrow \quad y^2=36\pm 12\sqrt6 +6\quad \rightarrow \quad y^2=42\pm 12\sqrt6

x^2=\bigg(\dfrac{6\pm \sqrt6}{2}\bigg)^2\quad \rightarrow \quad x^2=\dfrac{42\pm 12\sqrt6}{4}\quad \rightarrow \quad x^2=\dfrac{21\pm 6\sqrt6}{2}

                                                                                 

x^2+y^2=\dfrac{21\pm 6\sqrt6}{2}+42\pm 12\sqrt6\\\\\\.\qquad \quad = \dfrac{21\pm 6\sqrt6}{2}+\dfrac{84\pm 24\sqrt6}{2}\\\\\\. \qquad \quad = \dfrac{105\pm 18\sqrt6}{2}

5 0
3 years ago
I need help i will give the brainleness thing to whoever can answer
dimaraw [331]

Answer:

A=30cm, B=39cm

Step-by-step explanation:

They are an enlargement. Therefore, the sides are larger by the same proportion, otherwise, it wouldn't be an enlargement. Therefore, the bottom side of the big one is 3 times that of the small one. Therefore, the other sides are also 3 times, making A = 30cm and B = 39cm.

8 0
3 years ago
Read 2 more answers
The equation I wrote was F=9/5C+32 Rearrange the equation and solve for C
Trava [24]
F=9/5C+32
rewrite as 9/5C+32=F
9/5C+32-32=F-32
9/5C=F-32
(5*9/5)C=(F-32)*5
9C=5F-160
9/9C=(5F-160)/9
C=(5F-160)/9

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3 years ago
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ElenaW [278]
Thats all the question?
6 0
3 years ago
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