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oee [108]
3 years ago
8

Please Help

Mathematics
2 answers:
beks73 [17]3 years ago
7 0

Answer:

h=S/(2\pi r)

Step-by-step explanation:

we know that

The surface lateral area of a cylinder is equal to

S=2\pi rh

where

r is the radius of the base of the cylinder

h is the height of the cylinder

Solve for h-------> means isolate the variable h

so

Divide by 2\pi r both sides

S/(2\pi r)=2\pi rh/(2\pi r)

Simplify

h=S/(2\pi r)

Delicious77 [7]3 years ago
6 0
First you want to get h alone so you divide 2<span>π</span>r by s so your answer would be
h=S/2πr


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For questions 1 - 2, use the distributive property to find the product of the binomials.
igor_vitrenko [27]

Answer:

1).  2x^2-3x+20

2). 4x^2-14x+6

Step-by-step explanation:

1. (x-4)(2x+5)

FOIL:

(x)(2x)+5(x)-8(x)+(4)(5)

2x^2+5x-8x+20

=2x^2-3x+20

2. (4x-2)(x-3)

FOIL:

(4x)(x)-(3)(4x)-2(x)-2(-3)

4x^2-12x-2x+6

=4x^2-14x+6

5 0
2 years ago
Solve each single-step equation. Show your work. -15+n=-9
Serggg [28]

Answer:

6

Step-by-step explanation:

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3 0
2 years ago
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Which of the following are solutions to the equation below?
sladkih [1.3K]

Answer:

\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}

                  x=3,\:x=-\frac{1}{2}

Step-by-step explanation:

considering the equation

2x^2\:-\:4x\:-\:3\:=\:x

solving

2x^2\:-\:4x\:-\:3\:=\:x

2x^2-4x-3-x=x-x

2x^2-5x-3=0

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=2,\:b=-5,\:c=-3:\quad x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}

solving

x=\frac{-\left(-5\right)+\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}

x=\frac{5+\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}

x=\frac{5+\sqrt{49}}{2\cdot \:2}

x=\frac{5+7}{4}

x=3

also solving

x=\frac{-\left(-5\right)-\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}

x=\frac{5-\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}

x=\frac{5-\sqrt{49}}{4}

x=-\frac{2}{4}

x=-\frac{1}{2}

Therefore,

                 \mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}

                  x=3,\:x=-\frac{1}{2}

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Mice21 [21]
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n terms

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Sonbull [250]
X= 12 is your answer explanation?
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