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Firlakuza [10]
3 years ago
8

Find the area of the figure below?

Mathematics
1 answer:
inna [77]3 years ago
7 0

Answer:

i cant load the picture

Step-by-step explanation:

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Suppose you could climb to the top of the Great Pyramid of Giza in Egypt. Which path would be shorter, climbing a lateral edge o
valentinak56 [21]
Check the picture below.

if you go from the slant height, the path is shorter to the top, since the hypotenuse is the longest of all sides in a right-triangle.

that said, is shorter but is also steeper, so the lateral edge is longer but less steep and thus a bit simpler though longer, but we're only asked on which is shorter.

5 0
3 years ago
What is the slope of a line passing through (1,9) and (-3,16) ?
timurjin [86]

Answer:

slope= -7/4

Step-by-step explanation:

y2-y1/x2-x1

16-9/-3-1

7 0
3 years ago
From a window 20 feet above the ground, the angle of elevation to the top of a building across
Nikitich [7]

Answer: The answer is 381.85 feet.

Step-by-step explanation:  Given that a window is 20 feet above the ground. From there, the angle of elevation to the top of a building across  the street is 78°, and the angle of depression to the base of the same building is 15°. We are to calculate the height of the building across the street.

This situation is framed very nicely in the attached figure, where

BG = 20 feet, ∠AWB = 78°, ∠WAB = WBG = 15° and AH = height of the bulding across the street = ?

From the right-angled triangle WGB, we have

\dfrac{WG}{WB}=\tan 15^\circ\\\\\\\Rightarrow \dfrac{20}{b}=\tan 15^\circ\\\\\\\Rightarrow b=\dfrac{20}{\tan 15^\circ},

and from the right-angled triangle WAB, we have'

\dfrac{AB}{WB}=\tan 78^\circ\\\\\\\Rightarrow \dfrac{h}{b}=\tan 15^\circ\\\\\\\Rightarrow h=\tan 78^\circ\times\dfrac{20}{\tan 15^\circ}\\\\\\\Rightarrow h=361.85.

Therefore, AH = AB + BH = h + GB = 361.85+20 = 381.85 feet.

Thus, the height of the building across the street is 381.85 feet.

8 0
3 years ago
Could I get some help?
Liono4ka [1.6K]
Vertical angles 2 and 5,  1 and 6, 3 and 8, 7 and 4
linear pair  1 and 5, 5 and 6, 6 and 2, 2 and 1, and also 3 and 7, 7 and 8, 8 and 4, 4 and 3
congruent angles - all vertical pairs are congruent angles
 2 congruent to 5,  1 congruent to 6, 3 congruent to 8, 7 congruent to 4 
8 0
3 years ago
How do I factorise 2t^2 + 5t +2
alekssr [168]
2t^2+5t+2\\\\a=2;\ b=5;\ c=2\\\\\Delta=b^2-4ac;\ if\ \Delta > 0\ then\ t_1=\frac{-b-\sqrt\Delta}{2a}\ and\ t_2=\frac{-b+\sqrt\Delta}{2a}\\\\\Delta=5^2-4\cdot2\cdot2=25-16=9;\ \sqrt\Delta=\sqrt9=3\\\\t_1=\frac{-5-3}{2\cdot2}=\frac{-8}{4}=-2;\ t_2=\frac{-5+3}{2\cdot2}=\frac{-2}{4}=-\frac{1}{2}\\\\2t^2+5t+2=2(x+2)(x+\frac{1}{2})
5 0
3 years ago
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