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3 years ago

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You are building a model of a square pyramid. The side length of the base of the model is 30 cm. The height is 21 cm. What is th

e slant height of the model pyramid

1 answer:

Shalnov [3]3 years ago

6 0

Answer:

Step-by-step explanation:

find the area of the base (square)=900 sq. cm
find area of triangles=1260 sq. cm
total= 2160 sq. cm

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2 years ago

Read 2 more answers

A. x-int:-0.5,y-int:1<br>B. x-int:0.5,y-int:1<br>C. x-int:-0.5,y-int:-1<br>D. x-int:1,y-int:0.5

hodyreva [135]

The correct answer is: [D]: " <span>x-int : 1 , y-int: 0.5 " .
_____________________________________________________
Note:
_____________________________________________________
The "x-intercept" refers to the point(s) at which the the graph of a function (which is a "line", in this case) cross(es) the "y-axis".

In other words, what is (are) the point(s) of the graph at which "x = 0<span>" ?
</span>
By examining the graph, we see that when " x = 0" ; y is equal to: "1<span>" .
</span>
So; the "x-intercept" is at point: "(0, 1)" ; or, we can simply say that the
"x-intercept" is: "1" .
_________________________________________________________</span> Note:
_____________________________________________________
The "y-intercept" refers to the point(s) at which the the graph of a function (which is a line, in this case) cross(es) the "x-axis".

In other words, what is (are) the point(s) of the graph at which " y = 0 <span>" ?
</span>
By examining the graph, we see that when " y = 0 " ; x is equal to: "0.5<span>" .
</span>
So; the "x-intercept" is at point: "(0.5, 0)" ; or, we can simply say that the
"y-intercept" is: "0.5 " .<span>
______</span>_________________________________________________
This would correspond to:<span>
_______________________________________________________
Answer choice: [D]: </span>" x-int: 1 , y-int: 0.5 " .
_______________________________________________________
{that is; The "x-intercept" is: "0" ; and the "y-intercept" is: "0.5 ".} .
_______________________________________________________

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B

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

2 years ago