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sesenic [268]
3 years ago
6

What is the y-intercept in the equation LaTeX: y=-3x+7y = − 3 x + 7?

Mathematics
1 answer:
poizon [28]3 years ago
3 0
B answer is letter B
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Write a verbal sentence to represent 5x+7=18
Romashka-Z-Leto [24]
Five times x plus seven equals eighteen 
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3 years ago
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Use dimensional analysis to convert the quantity to the indicated units.
Eva8 [605]

Answer:

27 ft = 324 inches

Step-by-step explanation:

We want to use dimensional analysis to convert a quantity here

27 ft to inches

Mathematically;

12 inches = 1 ft

so;

x inches = 27 ft

So the number of inches will be 27 * 12 = 324 inches

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3 years ago
What expression is equivalent to 9(3x + 8)? *
Nat2105 [25]

Step-by-step explanation:

This is what it is equal to:

9×3x=27x

9×8=72

Therefore it will be:

27x + 72

hope it helps!

8 0
3 years ago
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Prove that sinA-sin3A+sin5A-sin7A/cosA-cos3A-cos5A+cos7A= cot2A
MAVERICK [17]
Write the left side of the given expression as N/D, where
N = sinA - sin3A + sin5A - sin7A
D = cosA - cos3A - cos5A + cos7A
Therefore we want to show that N/D = cot2A.

We shall use these identities:
sin x - sin y = 2cos((x+y)/2)*sin((x-y)/2)
cos x - cos y = -2sin((x+y)/2)*sin((x-y)2)

N = -(sin7A - sinA) + sin5A - sin3A
    = -2cos4A*sin3A + 2cos4A*sinA
    = 2cos4A(sinA - sin3A)
    = 2cos4A*2cos(2A)sin(-A)
    = -4cos4A*cos2A*sinA

D = cos7A + cosA - (cos5A + cos3A)
   = 2cos4A*cos3A - 2cos4A*cosA
   = 2cos4A(cos3A - cosA)
   = 2cos4A*(-2)sin2A*sinA
   = -4cos4A*sin2A*sinA

Therefore
N/D = [-4cos4A*cos2A*sinA]/[-4cos4A*sin2A*sinA]
       = cos2A/sin2A
      = cot2A

This verifies the identity.
4 0
3 years ago
Adventure Play sells an mp3 player for $18.00 and charges $3.25 per song. King Music sells a player for $23.00 and charges $2.00
Rus_ich [418]
An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or …show more content… 
An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or half-moon status because at least 50 percent of its surface was illuminated. In the following nights, the Moon displayed characteristics of waxing gibbous as the light continued to grow across the moon’s surface from right to left. The Moon was nearing closer to the full moon phase on November 14th as only a very small dark shadow was visible on the left side. 
The Moon takes 27.3 days (sidereal month) to complete its actual orbit around the Earth. Like the Sun, the Moon rises in the east and sets in the west each day. 
4 0
3 years ago
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