<img src="https://tex.z-dn.net/?f=g%28x%29%3Dx%5E3%2B2x-1" id="TexFormula1" title="g(x)=x^3+2x-1" alt="g(x)=x^3+2x-1" align="abs

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alexira [117]

2 years ago

middle" class="latex-formula">.

Mathematics

1 answer:

Anastasy [175]2 years ago

5 0

First you differentiate it. So:
x^3 + 2x - 1 = 3x^2 + 2
at point (2, 1) the x value is 2 so you substitute that into the differentiated equation.
you get 3(2)^2 + 2 = 14

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Evaluate each expression 10 (e+7) E=3

Nostrana [21]

Plug in e for 3.

10(3+7)

Use the distributive property. a(b+c)= ab+ac

10*3= 30, 10*7=70

30+70=100

The expression equals 100.

I hope this helps!
~kaikers

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3 years ago

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The graph of y = |x-6| is in the standard (x,y)

Airida [17]

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Answer:

F. Translation to the right 6 coordinate units

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3 years ago

If the radius of a circle with a area of 5.3 mm squared is multiplied by 5,

chubhunter [2.5K]

I'm guessing the question is meant to be "If the radius of a circle with a area of 5.3 mm squared is multiplied by 5, what will the new area be?", so i'll answer that.

Area of a circle =πr^2

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6 0

3 years ago

What is the multiplicative inverse of 5 in z11, z12, and z13? you can do a trial-and-error search using a calculator or a pc?

grin007 [14]

A multiplicative inverse of an integer a is an integer x such that the product ax is congruent to 1 with respect to the modulus m.

1. $Z_{11}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10}\}.$

Check:

$5\cdot 0=\overline{0};$
$5\cdot 1=\overline{5};$
$5\cdot 2=\overline{10};$
$5\cdot 3=15=\overline{4};$
$5\cdot 4=20=\overline{9};$
$5\cdot 5=25=\overline{3};$
$5\cdot 6=30=\overline{8};$
$5\cdot 7=35=\overline{2};$
$5\cdot 8=40=\overline{7};$
$5\cdot 9=45=\overline{1};$
$5\cdot 10=50=\overline{6}.$

The multiplicative inverse of 5 in $Z_{11}$ is 9.

2. $Z_{12}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10},\overline{11}\}.$

Check:

$5\cdot 0=\overline{0};$
$5\cdot 1=\overline{5};$
$5\cdot 2=\overline{10};$
$5\cdot 3=15=\overline{3};$
$5\cdot 4=20=\overline{8};$
$5\cdot 5=25=\overline{1};$
$5\cdot 6=30=\overline{6};$
$5\cdot 7=35=\overline{11};$
$5\cdot 8=40=\overline{4};$
$5\cdot 9=45=\overline{9};$
$5\cdot 10=50=\overline{2};$
$5\cdot 11=55=\overline{7}.$

The multiplicative inverse of 5 in $Z_{12}$ is 5.

3. $Z_{13}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10},\overline{11},\overline{12}\}.$