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3 years ago

What is the missing term? (10x – 4x2) – (7x + mc023-1.jpg) = 3x – 6x2

Mathematics

2 answers:

dimaraw [331]3 years ago

7 0

Answer:

$\text{The missing term is }2x^2$

Step-by-step explanation:

Given the expression

$(10x-4x^2)-(7x+missing\thinspace term)=3x-6x^2$

we have to find the missing term.

$(10x-4x^2)-(7x+missing\thinspace term)=3x-6x^2$

$10x-4x^2-7x-missing\thinspace term=3x-6x^2$

$\text{Adding }4x^2\text{from both sides}$

$10x-7x-missing\thinspace term=3x-6x^2+4x^2$

$3x-missing\thinspace term=3x-2x^2$

$missing\thinspace term=2x^2$

$\text{Hence, the missing term is }2x^2$

QveST [7]3 years ago

3 0

Answer:

The missing term is B. 2x^2

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Read 2 more answers

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