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PilotLPTM [1.2K]
3 years ago
11

50 is 50% of what number? Plz show work in explanation thanks. Again im in middle school i musta misclicked high school instead

Mathematics
1 answer:
erastovalidia [21]3 years ago
6 0

100

50 is half of 100 and 50% is half.

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Answer:

2.2.1 Area = 12.6 cm²

2.2.2 Area = 12 cm²

Step-by-step explanation:

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<u>2.2.1:</u>

The radius of the big circle is half the length of the square = 2.5 cm

The radius of the small circle is half of 3 cm = 1.5 cm

The area of the shaded portion is = Area of big circle - Area of small circle

π × 2.5²-π × 1.5² = 12.6 cm²

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To find length BD (the height of the parallelogram) we need to use the Pythagorean theorem:

BD = √(5²-3²) = 4

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Step-by-step explanation:

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MatroZZZ [7]

9514 1404 393

Answer:

  second step is wrong; x = 5

Step-by-step explanation:

The correct solution is ...

  3(x -3) +9 = 15

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2 years ago
f(x)=6x3−54x2−126x−8 is decreasing on the interval ( equation editorEquation Editor , equation editorEquation Editor ). It is in
worty [1.4K]

Answer:

The function f(x)=6x^3-54x^2-126x-8 is decreasing on the interval (-1,7) and it is increasing on the interval (-\infty, -1)\cup (7, \infty)

Step-by-step explanation:

To determine the intervals of increase and decrease of the function f(x)=6x^3-54x^2-126x-8, perform the following steps:

1. Differentiate the function

\frac{d}{dx}\left(6x^3-54x^2-126x-8\right)=\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\frac{d}{dx}\left(6x^3\right)-\frac{d}{dx}\left(54x^2\right)-\frac{d}{dx}\left(126x\right)-\frac{d}{dx}\left(8\right)\\\\f'(x)=18x^2-108x-126

2. Obtain the roots of the derivative, f'(x) = 0

\mathrm{Factor\:out\:common\:term\:}18:\quad 18\left(x^2-6x-7\right)\\\\\mathrm{Factor}\:x^2-6x-7:\quad \left(x+1\right)\left(x-7\right)\\\\18x^2-108x-126:\quad 18\left(x+1\right)\left(x-7\right)\\\\18x^2-108x-126=0\quad :\quad x=-1,\:x=7

3.  Form open intervals with the roots of the derivative and take a value from every interval and find the sign they have in the derivative.

If f'(x) > 0, f(x) is increasing.

If f'(x) < 0, f(x) is decreasing.

On the interval \left(-\infty, -1\right), take x = -2,

f'(-2)=18(-2)^2-108(-2)-126=162  f'(x) > 0 therefore f(x) is increasing

On the interval \left(-1,7), take x = 0,

f'(0)=18(0)^2-108(0)-126=-126  f'(x) < 0 therefore f(x) is decreasing

On the interval \left(7, \infty\right), take x = 10,

f'(10)=18(10)^2-108(10)-126=594  f'(x) > 0 therefore f(x) is increasing

The function f(x)=6x^3-54x^2-126x-8 is decreasing on the interval (-1,7) and it is increasing on the interval (-\infty, -1)\cup (7, \infty)

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3 years ago
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A=πr(r+h2+r2)

Step-by-step explanation:

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