Can someone help me?
Answers
Answer:
Problem 1:
The student is wrong, the lines have NO SOLUTION, infinite solutions, because the lines don't intersect in any areas.
Problem 2:
The student is correct, because the lines intersect at (-3,-2)
Step-by-step explanation:
Related Questions
Find the value of x.
What's the value of log2 (1/8) ?
The average speed of a plane is 560 kilometers per hour. How long does it take the plane to travel 140 kilometers?
Solve the system using multiplication for the linear combination method.6x – 3y = 3 –2x + 6y = 14 What is the solution to the system? (2, 1) (2, –3) (2, –1) (2, 3)
B.20/30
C.40/48
D.12/14
Answers
7x + 2y=66
I need a answer in words step by step please help!!
Answers
Answer:
x = 10
y = -2
Step-by-step explanation:
Elimination method:
6x - 5y = 70 -----------------(I)
7x + 2y = 66 -----------------(II)
Multiply equation (I) by 2 and equation (II) by 5 and add. So, y will be eliminated and you can now find the value of x
(I)*2 12x - 10y = 140
(II)*5 35x + 10y = 330 {Now add}
47x = 470
x = 470/47
x = 10
Plugin x = 10 in equation (II)
7*10 + 2y = 66
70 + 2y = 66
Subtract 70 from both sides
2y = 66 - 70
2y = -4
Divide both sides by 2
y = -4/2
y = - 2
Answers
Answer:7.5
Step-by-step explanation:
Answer:
The answer will be 8.5
Answers
Answer:
Sun is cm away from alpha century.
Sun is m.
Sun is 1.391016 time the size of an atom at this scale.
Step-by-step explanation:
Light year is a measure of distance. It is the distance light travels in an year.
Light year = km
So 4.4 light years = km
km
Lets scale this down to the level of
km
= km
Change the units to centimeters:
cm
= cm
= cm
Therefore on the new scale sun is cm away from alpha century.
Diameter of the sun is 1.391016 million km
Lets change Sun's diameter to the new scale:
km
=km
Lets change kilometers in to meters:
m
=m
Therefore, sun is m
and an atom is
Therefore the sun is 1.391016 time the size of an atom at this scale.
Final answer:
On a 1-to-10^19 scale, the distance from the Sun to the Alpha Centauri is about 44 cm. On this same scale, the Sun itself would have a diameter of about 150 picometers, which is larger than a typical atom.
Explanation:
The 1-to-10^19 scale means for every actual meter in space, we represent 10^19 meters on our model. The Alpha Centauri is 4.4 light-years away from the sun. Considering 1 light-year equals to approximately 9.46x10^15 meters, the real distance from the sun to Alpha Centauri is about 4.16x10^16 meters. So, on the scale, this is about 0.44 meters or 44 cm.
The Sun's real size, with a diameter of 1.5 million kilometers or 1.5x10^9 meters, is represented as 1.5x10^-10 meters or 150 picometers on the scale. This is much bigger than an actual atom, which has a diameter of 0.1 to 0.5 nanometers or 100 to 500 picometers. Hence, on this scale, the Sun would be larger than a typical atom.
Learn more about Scale Representation here:
#SPJ11
Answers
Answer:
To find the peak value of a sinusoidal alternating current given its RMS (Root Mean Square) value, you can use the relationship between peak value and RMS value for a sinusoidal waveform.
The relationship is given by:
Peak Value = RMS Value * √2
In this case, the RMS value is 4.78 amp. Plugging this into the equation, we get:
Peak Value = 4.78 amp * √2 ≈ 6.76 amp
Therefore, the correct answer is option b) 6.76 amp.
Answers
Answer:
The coordinates of D' are (1,-1)
Step-by-step explanation:
The point D in the figure has co-ordinates (2,-2) as shown in the figure.
The figure is dilated by a factor of 1/2
So, multiply the coordinates of D (2,-2) by 1/2
D' = (1/2*2, 1/2*-2)
D' = (1,-1)
So, the coordinates of D' are (1,-1)