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matrenka [14]
3 years ago
9

A cola container is in the shape of a right circular cylinder. The radius of the base is 4 centimeters, and the height is 10 cen

timeters. What is V, the volume of the container?
Mathematics
1 answer:
noname [10]3 years ago
5 0

Answer:

V=160\pi

Step-by-step explanation:

V=h\pi r^{2}

V=10*\pi 4^{2}

V=160\pi

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Which of the following are solutions to the equation below?
Ulleksa [173]

Answer:

see below

Step-by-step explanation:

4x^2-12x+9=5\\\\4x^2-12x+9-5=0\\\\(2x-3)(2x-3)=5\\\\(2x-3)^2=5\\\\2x-3=\pm\sqrt{5}\\\\2x=3\pm\sqrt{5}\\\\x=\frac{3\pm\sqrt{5}}{2}\\\\x_1=\frac{3+\sqrt{5}}{2}\\\\x_2=\frac{3-\sqrt{5}}{2}

choose yourself

I don't understand your options

3 0
3 years ago
Given f(x)=arctan(2x), find f’(x) in simplest form
Lostsunrise [7]

Answer:

Step-by-step explanation:

f(x)=tan^{-1} 2x\\let f(x)=y\\y=tan^{-1}2x\\tan y=2x\\diff. w.r.t.x\\sec^2y(\frac{dy}{dx} )=2\\\frac{dy}{dx} =\frac{2}{sec^2 y} =\frac{2}{1+tan^2y} =\frac{2}{1+(2x)^2} \\f'(x)=\frac{2}{1+4x^2}

7 0
3 years ago
Suppose the equation for line A is given by the equation y=--- 2. Ifline A and line B are perpendicular and the point (-4,1) lie
nignag [31]

Answer:

Not sure sorry

Step-by-step explanation:

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4 0
3 years ago
I NEED HELP PLS THIS IS DUE IN 3 HOURS
Mariulka [41]

Answer:

Part 1)  x^{2} -2x-2=(x-1-\sqrt{3})(x-1+\sqrt{3})

Part 2)  x^{2} -6x+4=(x-3-\sqrt{5})(x-3+\sqrt{5})

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

Part 1)

in this problem we have

x^{2} -2x-2=0

so

a=1\\b=-2\\c=-2

substitute in the formula

x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(1)(-2)}} {2(1)}\\\\x=\frac{2(+/-)\sqrt{12}} {2}\\\\x=\frac{2(+/-)2\sqrt{3}} {2}\\\\x_1=\frac{2(+)2\sqrt{3}} {2}=1+\sqrt{3}\\\\x_2=\frac{2(-)2\sqrt{3}} {2}=1-\sqrt{3}

therefore

x^{2} -2x-2=(x-(1+\sqrt{3}))(x-(1-\sqrt{3}))

x^{2} -2x-2=(x-1-\sqrt{3})(x-1+\sqrt{3})

Part 2)

in this problem we have

x^{2} -6x+4=0

so

a=1\\b=-6\\c=4

substitute in the formula

x=\frac{-(-6)(+/-)\sqrt{-6^{2}-4(1)(4)}} {2(1)}

x=\frac{6(+/-)\sqrt{20}} {2}

x=\frac{6(+/-)2\sqrt{5}} {2}

x_1=\frac{6(+)2\sqrt{5}}{2}=3+\sqrt{5}

x_2=\frac{6(-)2\sqrt{5}}{2}=3-\sqrt{5}

therefore

x^{2} -6x+4=(x-(3+\sqrt{5}))(x-(3-\sqrt{5}))

x^{2} -6x+4=(x-3-\sqrt{5})(x-3+\sqrt{5})

5 0
3 years ago
Plzzz Solve these 7 Questions ASAP !!<br> PLZZZZZZZZZZZ IT'S A REQUEST !!!
e-lub [12.9K]

Answer:

B is an answer.

6 0
4 years ago
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