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dalvyx [7]
3 years ago
9

A person who is 1.5 meters tall casts a shadow that is 8 meters long. The distance along the ground from the person (N) to the f

lagpole (G) is 32 meters. Find the height of the flagpole (FG) showing all calculations.
Mathematics
1 answer:
neonofarm [45]3 years ago
7 0
1.) 1.5m/8m= fg/32m
2.) 1.5 (32)= 8m * fg
3.) 1.5 (32)/8m= fg
4.) 6m = fg

Im pretty sure that's right
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11-Across plus 28-Down
Art [367]

Answer:

(11,-28)

Step-by-step explanation:

8 0
3 years ago
Solve for b. Express your answer as an integer or integers or in simplest radical form. -165= -3-6b^3
Fynjy0 [20]
I wanna help you but I don’t understand sorry
4 0
2 years ago
9) Nancy starts a race at the start line and she is running 3 meters per second. Juan starts the same race 3 meters ahead of Nan
Travka [436]

Given:

Nancy is running 3 meters per second.

Juan starts the same race 3 meters ahead of Nancy but he is going at 2 meters per second.

To find:

The equations for Nancy and Juan.

Solution:

Let x be the number of seconds.

Nancy is running 3 meters per second. So, the total distance covered by Nancy in the race is

y=3x

Juan starts the same race 3 meters ahead of Nancy but he is going at 2 meters per second. So, the total distance covered by Juan in the race is

y=3+2x

Therefore, the equations of Nancy and Juan are y=3x and y=3+2x respectively.

5 0
3 years ago
Please help me. It’s geometry
Verizon [17]

Option 3:

m∠ABC = 66°

Solution:

Given \overline {F A D} \| \overline {E H C} and ABH is a transversal line.

m∠FAB = 48° and m∠ECB = 18°

m∠ECB = m∠HCB = 18°

<u>Property of parallel lines: </u>

<em>If two parallel lines cut by a transversal, then the alternate interior angles are equal.</em>

m∠FAB = m∠BHC

48° = m∠BHC

m∠BHC = 48°

<u>Exterior angle of a triangle theorem: </u>

<em>An exterior angle of a triangle is equal to the sum of the opposite interior angles.</em>

m∠ABC = m∠BHC + m∠HCB

m∠ABC = 48° + 18°

m∠ABC = 66°

Option 3 is the correct answer.

4 0
3 years ago
Help me please. <br>Help me please. <br>Help me please.
Gnesinka [82]
A)
To be similar triangles have to have equal angles
triangle ZDB' 
1)angle Z=90 degrees
triangle B'CQ
1) angle C 90 degrees

angle A'B'Q=90
DB'Z+A'B'Q+CB'Q=180, straight angle
DB'Z+90+CB'Q=180
DB'Z+CB'Q=90

triangle ZDB'
DZB'+DB'Z=180-90=90

DB'Z+CB'Q=90
DZB'+DB'Z=90
DB'Z+CB'Q=DZB'+DB'Z
2)CB'Q=DZB' (these angles from two triangles ZDB' and B'CQ )
3)so,angles DB'Z and B'QC are going to be equal because of sum of three angles in triangles =180 degrees and 2 angles already equal.
so this triangles are similar by tree angles
b)
B'C:B'D=3:4
B'D:DZ=3:2
CQ-?
DC=AB=21
DC=B'C+B'D (3+4= 7 parts)
21/7=3
B'C=3*3=9
B'D=3*4=12
B'D:DZ=3:2
12:DZ=3:2
DZ=12*2/3=8
B'D:DZ=CQ:B'C
3:2=CQ:9
CQ=3*9/2=27/2

c)
BC=BQ+QC=B'Q+QC
BQ' can be found by pythagorean theorem



3 0
3 years ago
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