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Komok [63]
3 years ago
6

During a week in January in Jefferson City, MO, the daily high temperatures were -2˚F, 12˚F, 5˚F, -4˚F, 20˚F, 16˚F, and -12˚F. W

hich expression can be used to find the difference between the highest temperature and the lowest temperature of the week? A. 20 – 2 B. 20 – (-2) C. 20 – 12 D. 20 – (-12) PLEASE HELP QUICKLY!!!!
Mathematics
1 answer:
Alexeev081 [22]3 years ago
3 0

Answer:

50-20+5= 35

Step-by-step explanation:

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2 x e-4

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1. Consider the right triangle ABC given below.
lbvjy [14]
#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
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