When Scott woke up this morning, the temperature was 2°F. The temperature fell 5° by noon. Scott used this number line to model
the temperature drop.
What error did Scott make?
A. Scott should have started by moving from 0 to –5.
B. Scott should have started by moving from 0 to –2.
C. Scott started correctly by moving from 0 to 2, but then he should have moved 5 units to the left.
D. Scott started correctly by moving from 0 to 2, but then he should have moved 5 units to the right.
What error did Scott make?
A. Scott should have started by moving from 0 to –5.
B. Scott should have started by moving from 0 to –2.
C. Scott started correctly by moving from 0 to 2, but then he should have moved 5 units to the left.
D. Scott started correctly by moving from 0 to 2, but then he should have moved 5 units to the right.
2 answers:
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See the attached figure to better understand the problem
let
L-----> length side of the cuboid
W----> width side of the cuboid
H----> height of the cuboid
we know that
One edge of the cuboid has length 2 cm-----> <span>I'll assume it's L
so
L=2 cm
[volume of a cuboid]=L*W*H-----> 2*W*H
40=2*W*H------> 20=W*H-------> H=20/W------> equation 1
[surface area of a cuboid]=2*[L*W+L*H+W*H]----->2*[2*W+2*H+W*H]
100=</span>2*[2*W+2*H+W*H]---> 50=2*W+2*H+W*H-----> equation 2
substitute 1 in 2
50=2*W+2*[20/W]+W*[20/W]----> 50=2w+(40/W)+20
multiply by W all expresion
50W=2W²+40+20W------> 2W²-30W+40=0
using a graph tool------> to resolve the second order equation
see the attached figure
the solutions are
13.52 cm x 1.48 cm
so the dimensions of the cuboid are
2 cm x 13.52 cm x 1.48 cm
or
2 cm x 1.48 cm x 13.52 cm
<span>Find the length of a diagonal of the cuboid
</span>diagonal=√[(W²+L²+H²)]------> √[(1.48²+2²+13.52²)]-----> 13.75 cm
the answer is
the length of a diagonal of the cuboid is 13.75 cm
let
L-----> length side of the cuboid
W----> width side of the cuboid
H----> height of the cuboid
we know that
One edge of the cuboid has length 2 cm-----> <span>I'll assume it's L
so
L=2 cm
[volume of a cuboid]=L*W*H-----> 2*W*H
40=2*W*H------> 20=W*H-------> H=20/W------> equation 1
[surface area of a cuboid]=2*[L*W+L*H+W*H]----->2*[2*W+2*H+W*H]
100=</span>2*[2*W+2*H+W*H]---> 50=2*W+2*H+W*H-----> equation 2
substitute 1 in 2
50=2*W+2*[20/W]+W*[20/W]----> 50=2w+(40/W)+20
multiply by W all expresion
50W=2W²+40+20W------> 2W²-30W+40=0
using a graph tool------> to resolve the second order equation
see the attached figure
the solutions are
13.52 cm x 1.48 cm
so the dimensions of the cuboid are
2 cm x 13.52 cm x 1.48 cm
or
2 cm x 1.48 cm x 13.52 cm
<span>Find the length of a diagonal of the cuboid
</span>diagonal=√[(W²+L²+H²)]------> √[(1.48²+2²+13.52²)]-----> 13.75 cm
the answer is
the length of a diagonal of the cuboid is 13.75 cm
Answer:
Step-by-step explanation:
If your mind works better in degrees than it does radians, as does mine, it will be beneficial to us to see what degree measure this angle is. Convert it to degrees using the fact that 180° = π:
degrees
Now that we know that, we can plot that angle in a coordinate plane. The terminal side of the angle lands in quadrant 2. To find the reference angle, we subtract 120 from 180 and get that the reference angle is 60 degrees, which is the same as π/3.