Answer:
The height of the building is approximately <u>73 m</u>.
Step-by-step explanation:
The triangle EFG is shown below.
Given:
Length of shadow of building is, 
Angle of inclination of Sun with the vertical is 25°.
Therefore, the angle of inclination of Sun with the horizontal is given as:
Now, in triangle EFG as shown below, we use the tangent ratio to determine the height of building EF. Therefore,
Therefore, the height of the building is approximately 73 m.
Y0ou find a common denominatior, in this case is 15.... then multiply whatever u had to get the denominator 15 times the top number, so 3/5 + 10/15 = 13/15
The graph at option 1 shows the given inequality y < x² + 1. The domain and range of the given inequality is {x: x ∈ (-∞, ∞)} and {y: y ∈ [1, ∞)}.
<h3>How to graph an inequality?</h3>
The steps to graph an inequality equation are:
- Solve for the variable y in the given equation
- Graph the boundary line for the inequality
- Shade the region that satisfies the inequality.
<h3>Calculation:</h3>
The given inequality is y < x² + 1
Finding points to graph the boundary line by taking y = x² + 1:
When x = -2,
y = (-2)² + 1 = 4 + 1 = 5
⇒ (-2, 5)
When x = -1,
y = (-1)² + 1 = 2
⇒ (-1, 2)
When x = 0,
y = (0)² + 1 = 1
⇒ (0, 1)
When x = 1,
y = (1)² + 1 = 2
⇒ (1, 2)
When x = 2,
y = (2)² + 1 = 5
⇒ (2, 5)
Plotting these points in the graph forms an upward-facing parabola.
So, all the points above the vertex of the parabola satisfy the given inequality. Thus, that part is shaded.
From this, the graph at option 1 is the required graph for the inequality y < x² + 1. The boundary line is dashed since the inequality symbol is " < ".
Learn more about graphing inequalities here:
brainly.com/question/371134
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Answer with Step-by-step explanation:
The given differential euation is
![\frac{dy}{dx}=(y-5)(y+5)\\\\\frac{dy}{(y-5)(y+5)}=dx\\\\(\frac{A}{y-5}+\frac{B}{y+5})dy=dx\\\\\frac{1}{100}\cdot (\frac{10}{y-5}-\frac{10}{y+5})dy=dx\\\\\frac{1}{100}\cdot \int (\frac{10}{y-5}-\frac{10}{y+5})dy=\int dx\\\\10[ln(y-5)-ln(y+5)]=100x+10c\\\\ln(\frac{y-5}{y+5})=10x+c\\\\\frac{y-5}{y+5}=ke^{10x}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%3D%28y-5%29%28y%2B5%29%5C%5C%5C%5C%5Cfrac%7Bdy%7D%7B%28y-5%29%28y%2B5%29%7D%3Ddx%5C%5C%5C%5C%28%5Cfrac%7BA%7D%7By-5%7D%2B%5Cfrac%7BB%7D%7By%2B5%7D%29dy%3Ddx%5C%5C%5C%5C%5Cfrac%7B1%7D%7B100%7D%5Ccdot%20%28%5Cfrac%7B10%7D%7By-5%7D-%5Cfrac%7B10%7D%7By%2B5%7D%29dy%3Ddx%5C%5C%5C%5C%5Cfrac%7B1%7D%7B100%7D%5Ccdot%20%5Cint%20%28%5Cfrac%7B10%7D%7By-5%7D-%5Cfrac%7B10%7D%7By%2B5%7D%29dy%3D%5Cint%20dx%5C%5C%5C%5C10%5Bln%28y-5%29-ln%28y%2B5%29%5D%3D100x%2B10c%5C%5C%5C%5Cln%28%5Cfrac%7By-5%7D%7By%2B5%7D%29%3D10x%2Bc%5C%5C%5C%5C%5Cfrac%7By-5%7D%7By%2B5%7D%3Dke%5E%7B10x%7D)
where
'k' is constant of integration whose value is obtained by the given condition that y(2)=0\\
Thus the solution of the differential becomes
