Answer:
Part 1) [tex]c=\sqrt{146}\ cm[/tex]
Part 2) [tex]c=8\sqrt{2}\ in[/tex]
Part 3) The diagonal of rectangle is [tex]d=3\sqrt{29}\ cm[/tex]
Part 4) The length of the diagonal of computer monitor is [tex]d=15\ in[/tex]
Part 5) The ramp is [tex]10.59\ ft[/tex] long
Part 6) The distance between their houses is [tex]=4\sqrt{2}\ mi[/tex]
Part 7) [tex]234.31\ mi[/tex]
Part 8) 120 feet of wire is required
Step-by-step explanation:
Part 1) we know that
To find the length of the hypotenuse in a right triangle apply the Pythagorean Theorem
[tex]c^2=a^2+b^2[/tex]
where
c is the hypotenuse (the greater side)
a and b are the legs
we have
[tex]a=11\ cm\\b=5\ cm[/tex]
substitute
[tex]c^2=11^2+5^2[/tex]
[tex]c^2=146[/tex]
[tex]c=\sqrt{146}\ cm[/tex]
Part 2) we know that
To find the length of the hypotenuse in a right triangle apply the Pythagorean Theorem
[tex]c^2=a^2+b^2[/tex]
where
c is the hypotenuse (the greater side)
a and b are the legs
we have
[tex]a=8\ in\\b=8\ in[/tex]
substitute
[tex]c^2=8^2+8^2[/tex]
[tex]c^2=128[/tex]
[tex]c=\sqrt{128}\ in[/tex]
simplify
[tex]c=8\sqrt{2}\ in[/tex]
Part 3) we know that
To find the length of the diagonal in a rectangle apply the Pythagorean Theorem
[tex]d^2=b^2+h^2[/tex]
where
d is the diagonal of rectangle
b and h are the base and the height of rectangle
we have
[tex]b=15\ cm\\hb=6\ cm[/tex]
substitute
[tex]d^2=15^2+6^2[/tex]
[tex]d^2=261[/tex]
[tex]d=\sqrt{261}\ cm[/tex]
simplify
[tex]d=3\sqrt{29}\ cm[/tex]
Part 4) we know that
To find the length of the diagonal of a computer monitor apply the Pythagorean Theorem
[tex]d^2=w^2+h^2[/tex]
where
d is the diagonal of computer monitor
w and h are the wide and the high of computer monitor
we have
[tex]w=12\ in\\h=9\ in[/tex]
substitute
[tex]d^2=12^2+9^2[/tex]
[tex]d^2=225[/tex]
[tex]d=\sqrt{225}\ in[/tex]
simplify
[tex]d=15\ in[/tex]
Part 5) we know that
To find out the length of the ramp apply the Pythagorean Theorem
[tex]L^2=x^2+y^2[/tex]
where
L is the length of the ramp
x is the horizontal distance of the ramp
y is the vertical distance of the ramp
we have
[tex]x=10\ ft\\y=3.5\ ft[/tex]
substitute
[tex]L^2=10^2+3.5y^2[/tex]
[tex]L^2=112.25[/tex]
[tex]L=10.59\ ft[/tex]
Part 6) we know that
To find the distance between their houses apply the Pythagorean Theorem
[tex]c^2=a^2+b^2[/tex]
where
c is the hypotenuse (distance between their houses)
a and b are the legs
we have
[tex]a=4\ mi\\b=4\ mi[/tex]
substitute
[tex]c^2=4^2+4^2[/tex]
[tex]c^2=32[/tex]
[tex]c=4\sqrt{2}\ mi[/tex]
Part 7) we know that
The speed is equal to divide the distance by the time
[tex]speed=distance/time[/tex]
so
the distance is equal to multiply the speed by the time
[tex]distance=speed*time[/tex]
First train
speed=60 mph
time=3 hours
distance=60(3)=180 miles
Second train
speed=50 mph
time=3 hours
distance=50(3)=150 miles
To find out how far apart are the trains at the end of 3 hours apply the Pythagorean Theorem
[tex]c^2=a^2+b^2[/tex]
where
c is the hypotenuse (distance between the trains)
a and b are the legs
we have
[tex]a=180\ mi\\b=150\ mi[/tex]
substitute
[tex]c^2=180^2+150^2[/tex]
[tex]c^2=54,900[/tex]
[tex]c=234.31\ mi[/tex]
Part 8) we know that
For one tree is needed three wire
To find out the length of one wire apply the Pythagorean Theorem
[tex]c^2=a^2+b^2[/tex]
where
c is the hypotenuse (length of one wire)
a and b are the distance above the ground and the distance from the base
we have
[tex]a=3\ ft\\b=4\ ft[/tex]
substitute
[tex]c^2=3^2+4^2[/tex]
[tex]c^2=25[/tex]
[tex]c=5\ ft[/tex]
so
For one tree is required ----> (5)3=15 ft
therefore
For 8 trees is required
Multiply by 8
15(8)=120 ft
