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2 years ago

High-definition (HD) televisions today have a 16 : 9 aspect ratio (width to height). The advertised screen size is equal

Mathematics

2 answers:

madam [21]2 years ago

7 0

Answer:

B)68 by 40

Step-by-step explanation:

First we multiply this ratio by any whole number 16 : 9 . At least to get close to the option.

So I multiplied by 4

Giving me 64:36

So we remember that there is a padding of on both sides of the length and width.

This padding is 2 inch each

It's going to be

= 64+(2+2):36+(2+2)

= 68:40

sp2606 [1]2 years ago

3 0

Answer:

The most correct option is;

B) 68 by 40

Step-by-step explanation:

Here we have that the advertised screen size = Screen diagonal

Size of the HD TV = 75-inch

Therefore, diagonal of HD TV = 75-inch

Where the width of the HD TV = w

The height of the HD TV = h

h² + w² = 75²

w/h = 16/9

w = 16/9×h

∴ h² + (16/9×h)² = 75²

337/81×h² = 75²

h = 36.8 in.

Therefore, w = 16/9×36.8 = 65.4 in.

Hence when the thickness of the Styrofoam cup is added, we have;

2 inches on both sides gives 4 inches to be added to both height and width

Which gives,height = 36.8 + 4 = 40.8 in.

Width = 65.4 + 4 = 69.4 in.

The smallest possible dimensions is approximately

69.4 in. by 40.8 in.

Hence, the most correct option is B) 68 by 40.

You might be interested in

A firkin is an old British unit of volume equal to 9 gallons. How many cubic meters are there in 6.00 firkins?

Rainbow [258]

Answer:

0.245489 cubic meters.

Step-by-step explanation:

It is given that a firkin is an old British unit of volume equal to 9 gallons.

1 firkin = 9 gallons

6 firkins = 54 gallons

We need convert 6.00 firkins to cubic meters.

We know that,

1 gallon = 0.00454609 cubic meters

54 gallon = 0.24548886 cubic meters

6 firkins = 0.245489 cubic meters.

Therefore, 6.00 firkins is equivalent to 0.245489 cubic meters.

7 0

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