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shtirl [24]
3 years ago
7

To solve conditional statements

Mathematics
1 answer:
vesna_86 [32]3 years ago
5 0
When we prouieusly discussed inductive reasoning web based our reasoning an examples and on data from earlier events
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a solid consists of cylinder surmounted by a right circular cone. The height of the cone is 2h. If the volume of the solid is 3
Mademuasel [1]

Answer:

height of cylinder = 4/3 h

Step-by-step explanation:

The solid has a cylinder surmounted with a cone .Therefore, the volume of the solid is the sum of the cone and the cylinder.

volume of the solid = volume of cylinder + volume of cone

volume of the solid = πr²h + 1/3πr²h

let

height of the cylinder = H

recall

the height of the cone = 2h

volume of the solid = πr²h + 1/3πr²h

3(1/3πr²2h) = πr²H + 1/3πr²2h

2πr²h = πr²H + 2/3 πr²h

πr²(2h) = πr²(H + 2/3 h)

divide both sides by πr²

2h = H + 2/3 h

2h - 2/3h = H

H = 6h - 2h/3

H = 4/3 h

height of cylinder = 4/3 h

8 0
3 years ago
Convert 29.5% to a decimal
cricket20 [7]
Whatever% of anything is just (whatever/100) * anything.

thus 29.5% of something, is just (29.5/100) * something, and the decimal form of 29.5% is just 29.5/100 or the quotient of 29.5÷100.
7 0
3 years ago
Y=f(x)=(1/3)^x find f(x) when x=2
Sladkaya [172]

Answer:

f(2)=1/9

Step-by-step explanation:

f(x)=(1/3)^x

f(2)=(1/3)^(2)

f(2)=1/9

6 0
3 years ago
Please help me with this
Ne4ueva [31]

Answer:

Option c

Step-by-step explanation:

A system of equations are given to us. And we need to solve them . The given system is

\begin{cases} y = 2x - 3\dots 1 \\ y = x^2 - 3\dots 2 \end{cases}

We numbered the equations here . Now put the value of equation 1 in equation 2 that is substituting y = 2x - 3 in eq. 2 .

\implies y = x^2 - 3 \\\\\implies 2x - 3 = x^2 - 3 \\\\\implies x^2 - 2x = 0 \\\\\implies x(x-2) = 0 \\\\\implies\red{ x = 0 , 2 }

We got two values of x as 0 & 2 . Alternatively substituting these values we have ,

\implies y = 2 x - 3 \\\\\implies y = 2(0)-3 \qquad or \qquad y = 2(2)-3 \\\\\implies y = 0-3 \qquad or \qquad 4 - 3 \\\\\implies \red{ y = -3 , 1 }

Thefore the required answer is ,

\red{Option\:c} \begin{cases} (0,-3) \\ (2,1) \end{cases}

4 0
3 years ago
Read 2 more answers
15 Question Help
kari74 [83]

Answer:

pppp

Step-by-step explanation:

ppppppppppppppppppppppppppp

6 0
3 years ago
Read 2 more answers
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