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Marizza181 [45]
3 years ago
11

find the width, ans height and area of 26'' TV screen that has an aspect ratio of 4:3. round to the nearest whole number.

Mathematics
1 answer:
Rasek [7]3 years ago
6 0

erform a simple calculation to match the screen size of a standard TV to that of a widescreen TV. If you currently have a 4:3 TV and you want to continue watching 4:3 on a widescreen TV, multiply the diagonal length of the older TV model by 1.22. The result would be the diagonal screen size that the widescreen TV would have to be to match the old model.

<span>Say you have a 40 inch (102 cm) TV with a 4:3 aspect ratio, but you're thinking about upgrading and you don't want your screen size to get smaller. You'd need to get at least a 50 inch (127 cm) screen to view in 4:3 without your picture getting smaller. That's because 1.22 x 40 = 49. Since 49 inch TVs are generally not made, you'd need to go up to 50 inches (127 cm).</span>
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Evaluate each trigonometric equation for θ.
liberstina [14]

Answer:

sinФ = 8/17

cosФ = 15/17

tan Ф = 8/15

csc Ф = 17/8

sec Ф = 17/15

cot Ф = 15/8

Step-by-step explanation:

Let us revise the trigonometry functions

  • Sin(x) = opposite/hypotenuse
  • Cos(x) = adjacent/hypoteouse
  • Tan(x) = opposite/adjacent
  • Csc(x) = hypotenuse/opposit
  • Sec(x) = hypotenues/adjacent
  • Cot(x) = adjacent/opposite

In the given figure

The opposite side to angle Ф = 8

The adjacent side to angle Ф = 15

Find the hypotenuse using Pythagoras' theorem

Hypotenuse = \sqrt{8^{2}+15^{2}  }

Hypotenuse = \sqrt{64+225}

Hypotenuse = \sqrt{289}

Hypotenuse = 17

Let us use the rules above to find the trigonometry functions

sinФ = 8/17

cosФ = 15/17

tan Ф = 8/15

csc Ф = 17/8

sec Ф = 17/15

cot Ф = 15/8

4 0
3 years ago
The sum of 4 times a number and 7 more than the number is 80
marusya05 [52]

Answer:

(4×x) 7> 80  :)

Step-by-step explanation:

4 0
2 years ago
What is the length of line tAH
TEA [102]
It is 27 or less for tah
8 0
3 years ago
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Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in t
nataly862011 [7]

Let (<em>x</em>, <em>y</em>, <em>z</em>) be a point on the plane in the first octant. The box formed by this point has volume <em>xyz</em>, and you want to maximize this subject to the equation of the plane.

Use the method of Lagrange multipliers: the Lagrangian is

<em>L</em>(<em>x</em>, <em>y</em>, <em>z</em>) = <em>xyz</em> - <em>λ</em> (<em>x</em> + 2<em>y</em> + 3<em>z</em> - 6)

Find its critical points:

∂<em>L</em>/∂<em>x</em> = <em>yz</em> - <em>λ</em> = 0

∂<em>L</em>/∂<em>y</em> = <em>xz</em> - 2<em>λ</em> = 0

∂<em>L</em>/∂<em>z</em> = <em>xy</em> - 3<em>λ</em> = 0

∂<em>L</em>/∂<em>λ</em> = -(<em>x</em> + 2<em>y</em> + 3<em>z</em> - 6) = 0

Solving the first three equations for <em>λ</em> gives

<em>λ</em> = <em>yz</em> = <em>xz</em>/2 = <em>xy</em>/3

Solve these equations for <em>y</em> and <em>z</em> :

• <em>yz</em> = <em>xz</em>/2   =>   <em>y</em> = <em>x</em>/2   =>   2<em>y</em> = <em>x</em>

• <em>yz</em> = <em>xy</em>/3   =>   <em>z</em> = <em>x</em>/3   =>   3<em>z</em> = <em>x</em>

Substitute these solutions into the last equation and solve for <em>x</em>, then again for <em>y</em> and <em>z</em> :

<em>x</em> + 2<em>y</em> + 3<em>z</em> - 6 = 3<em>x</em> - 6 = 0   =>   3<em>x</em> = 6   =>   <em>x</em> = 2, <em>y</em> = 1, <em>z</em> = 2/3

At this critical point, the maximum volume is

<em>xyz</em> = 2*1*2/3 = 4/3

4 0
3 years ago
Fill in the missing number to name a fraction <br> equivalent to 6/8
Nataly [62]
3/4 would be a fraction, equivalent to 6/8.
4 0
4 years ago
Read 2 more answers
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