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3 years ago

Can someone help me please.

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

8 0

Answer:

Lena's Method: 280
Jackson's Method: 280

Step-by-step explanation:

Lena's Method:

8 + 12 = 20
Plug 20 in: 14(20)
14(20) = 14 × 20 = 280

Jackson's Method:

14 × 8 = 112
14 × 12 = 168
Re-write the expression: 112 + 168
112 + 168 = 280

I hope this helps!

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Given a soda can with a volume of 15 and a diameter of 2, what is the volume of a cone that fits perfectly inside the soda can?

Naddika [18.5K]

Check the picture below.

so, a cone, that fits inside the cylinder, will have the same height "h" and radius "r". Thus

$\bf \textit{volume of a cylinder}\\\\
V=\pi r^2 h\qquad \boxed{\pi r^2 h}=15
\\\\\\
\textit{volume of a cone}\\\\
V=\cfrac{\pi r^2 h}{3}\implies V=\cfrac{\boxed{15}}{3}$

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