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4 years ago

The point of a square pyramid is cut off, making each lateral face of the pyramid a trapezoid with the dimensions shown.

Mathematics

2 answers:

omeli [17]4 years ago

8 0

Answer:

The area of one trapezoidal face of the figure is 2 square inches

Step-by-step explanation:

<u><em>The complete question is</em></u>

The point of a square pyramid is cut off, making each lateral face of the pyramid a trapezoid with the dimensions shown. What is the area of one trapezoidal face of the figure?

we know that

The area of a trapezoid is given by the formula

$A=\frac{1}{2}(b_1+b_2)h$

where

b_1 and b-2 are the parallel sides

h is the height of the trapezoid (perpendicular distance between the parallel sides)

we have

$h=1\ in\\b_1=1\ in\\b_2=3\ in$

substitute the given values in the formula

$A=\frac{1}{2}(1+3)(1)$

$A=2\ in^2$

Irina-Kira [14]4 years ago

3 0

Answer:

2 square inches

Step-by-step explanation:

Hope this helps!

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inysia [295]

$\large \bigstar \frak{ } \large\underline{\sf{Solution-}}$

Consider, LHS

$\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} \\ \\ \text{So, using this identity, we get} \\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered} \\$

So, using this identity, we get

$\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}$

<h2>Hence,</h2>

$\begin{gathered} \\ \rm\implies \:\boxed{\sf{ \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}$

$\rule{190pt}{2pt}$

5 0

3 years ago

Use benchmarks <br> 3\5 + 3\17

nadya68 [22]

Answer:

66/85

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Calculati:

kramer

Answer:

Step-by-step explanation:

[(1/3 – 1/9)² + (1/3)³] : (1/3)⁴

Next, we shall simplify (1/3 – 1/9)². This is illustrated below:

(1/3 – 1/9)² = ((3 – 1)/9)² = (2/9)²

[(1/3 – 1/9)² + (1/3)³] : (1/3)⁴

= [(2/9)² + (1/3)³] : (1/3)⁴

= [4/81 + 1/27] : 1/81

= [(4 + 3)/81 ] : 1/81

= 7/81 : 1/81

= 7/81 ÷ 1/81

Invert

= 7/81 × 81/1

= 7

5 0

3 years ago

(7y^6z^4-4y^2)(-6yz^4)

rosijanka [135]

I dont lnow the awnser of that but good luck on that!

7 0

4 years ago

According to the Oxnard College Student Success Committee report in the previous year, we believe that 18% of students struggle

gladu [14]

Answer:

A sample size of 392 is required.

Step-by-step explanation:

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