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9966 [12]

3 years ago

s="latex-formula">

Mathematics

2 answers:

Jobisdone [24]3 years ago

4 0

X=-15,-15+y Y=15

Hope this helps

Have a great day/night

Stels [109]3 years ago

4 0

X is a ghost number which means it is considered alone so it’s just 15

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Could somebody help me please

pantera1 [17]

Answer:

its 7

Step-by-step explanation:

7 0

3 years ago

The length of a shadow of a tree is 150 feet when the angle of elevation of the sun is 39°. Approximate the height of the tree.

fomenos

Answer:

The height of tree is approximately 121.5 feet.

Step-by-step explanation:

We are given following in question:

Length of shadow = 150 feet

Angle of elevation of the sun =

$\theta = 39^\circ$

The tree and the shadow forms a right angled triangle.

Thus, with the help of trigonometric relation, we can write:

$\tan \theta = \dfrac{\text{Height of tree}}{\text{Length of shadow}}$

Let x feet be the height of tree.

Putting all the values, we get,

$\tan 39^\circ = \dfrac{x}{150}\\\\0.80978 = \dfrac{x}{150}\\\\\Rightarrow x = 150\times 0.80978\\\Rightarrow x = 121.467 \approx 121.5$

Thus, the height of tree is approximately 121.5 feet.

6 0

4 years ago

Given the graph of f(x) above, what is f(-4) *

earnstyle [38]

Answer:

Step-by-step explanation:

go to -4 on the x axis, then see which point the line is on, on the y axis.

3 0

3 years ago

What is 13i⋅(1−5i)? like i have no idea

madam [21]

Answer:

65 + 13i

Step-by-step explanation:

reminder that i² = - 1

13i(1 - 5i) ← multiply each term in the parenthesis by 13i

= 13i - 65i²

= 13i - 65(- 1)

= 13i + 65

= 65 + 13i ← in standard form

4 0

3 years ago

Use the squared identities to simplify 2sin^2xcos^2x

Rainbow [258]

$\bf sin^2({{ \theta}})=\cfrac{1-cos(2{{ \theta}})}{2}\qquad \qquad 

cos^2({{ \theta}})=\cfrac{1+cos(2{{ \theta}})}{2}\\
-----------------------------\\
thus
\\\\
2sin^2(x)cos^2(x)\implies \boxed{2}\cdot \cfrac{1-cos(2{{ x}})}{\boxed{2}}\cdot \cfrac{1+cos(2{{ x}})}{2}
\\\\
\cfrac{1-cos(2{{ x}})}{1}\cdot \cfrac{1+cos(2{{ x}})}{2}\implies 
\cfrac{[1-cos(2x)][1+cos(2x)]}{2}\\
-----------------------------\\
\textit{now recall}\implies \textit{difference of squares}
\\ \quad \\
$

$\bf (a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\
-----------------------------\\
thus
\\\\
\cfrac{[1-cos(2x)][1+cos(2x)]}{2}\implies \cfrac{1^2-cos^2(x)}{2}
\\\\
 \cfrac{1-cos^2(x)}{2}\impliedby 
\begin{array}{llll}
\textit{pretty sure you know what that numerator is}\\
\textit{check your pythagorean identities}
\end{array}$

7 0

3 years ago

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