The lateral area of the cube is 36 sq units
<u>Explanation:</u>
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Given:
Perimeter of the base of the cube = 12 units
Base of the cube has 4 sides
So, the perimeter of the base = 4a
where,
a is the length of one side
Thus,
4a = 12 units
a = 3 units
Lateral surface area of the cube = 4a²
A = 4 X (3)²
A = 36 sq units
Therefore, the lateral area of the cube is 36 sq units
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Find Slope :
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3y = -4x + 2
y = -4/3x + 2/3
Slope = - 4/3
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Answer: -4/3
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5^2 + 12^2 = c^2
25 + 144 = c^2
169 = c^2
take the square root of both sides
+ - 13 = c
Answer:
-2/3
Step-by-step explanation:
The slope of a line can be represented as 
where (x1, y1) and (x2, y2) are points on the line. We can substitute the points given, (-3, 5) and (6, -1), to calculate the slope:
#1)
A) b = 10.57
B) a = 22.66; the different methods are shown below.
#2)
A) Let a = the side opposite the 15° angle; a = 1.35.
Let B = the angle opposite the side marked 4; m∠B = 50.07°.
Let C = the angle opposite the side marked 3; m∠C = 114.93°.
B) b = 10.77
m∠A = 83°
a = 15.11
Explanation
#1)
A) We know that the sine ratio is opposite/hypotenuse. The side opposite the 25° angle is b, and the hypotenuse is 25:
sin 25 = b/25
Multiply both sides by 25:
25*sin 25 = (b/25)*25
25*sin 25 = b
10.57 = b
B) The first way we can find a is using the Pythagorean theorem. In Part A above, we found the length of b, the other leg of the triangle, and we know the measure of the hypotenuse:
a²+(10.57)² = 25²
a²+111.7249 = 625
Subtract 111.7249 from both sides:
a²+111.7249 - 111.7249 = 625 - 111.7249
a² = 513.2751
Take the square root of both sides:
√a² = √513.2751
a = 22.66
The second way is using the cosine ratio, adjacent/hypotenuse. Side a is adjacent to the 25° angle, and the hypotenuse is 25:
cos 25 = a/25
Multiply both sides by 25:
25*cos 25 = (a/25)*25
25*cos 25 = a
22.66 = a
The third way is using the other angle. First, find the measure of angle A by subtracting the other two angles from 180:
m∠A = 180-(90+25) = 180-115 = 65°
Side a is opposite ∠A; opposite/hypotenuse is the sine ratio:
a/25 = sin 65
Multiply both sides by 25:
(a/25)*25 = 25*sin 65
a = 25*sin 65
a = 22.66
#2)
A) Let side a be the one across from the 15° angle. This would make the 15° angle ∠A. We will define b as the side marked 4 and c as the side marked 3. We will use the law of cosines:
a² = b²+c²-2bc cos A
a² = 4²+3²-2(4)(3)cos 15
a² = 16+9-24cos 15
a² = 25-24cos 15
a² = 1.82
Take the square root of both sides:
√a² = √1.82
a = 1.35
Use the law of sines to find m∠B:
sin A/a = sin B/b
sin 15/1.35 = sin B/4
Cross multiply:
4*sin 15 = 1.35*sin B
Divide both sides by 1.35:
(4*sin 15)/1.35 = (1.35*sin B)/1.35
(4*sin 15)/1.35 = sin B
Take the inverse sine of both sides:
sin⁻¹((4*sin 15)/1.35) = sin⁻¹(sin B)
50.07 = B
Subtract both known angles from 180 to find m∠C:
180-(15+50.07) = 180-65.07 = 114.93°
B) Use the law of sines to find side b:
sin C/c = sin B/b
sin 52/12 = sin 45/b
Cross multiply:
b*sin 52 = 12*sin 45
Divide both sides by sin 52:
(b*sin 52)/(sin 52) = (12*sin 45)/(sin 52)
b = 10.77
Find m∠A by subtracting both known angles from 180:
180-(52+45) = 180-97 = 83°
Use the law of sines to find side a:
sin C/c = sin A/a
sin 52/12 = sin 83/a
Cross multiply:
a*sin 52 = 12*sin 83
Divide both sides by sin 52:
(a*sin 52)/(sin 52) = (12*sin 83)/(sin 52)
a = 15.11
