−2+7+(−4) pls help plssssssssssssssss
1 answer:
You might be interested in
Answer:
#1) y = -4x + 1; #2) parallel; #3) y = -1/2x + 5/2; #4) y = 2x + 4; #5) Never
Step-by-step explanation:
#1) We want a line that passes through (1, -3) and is parallel to y = 2-4(x-1). First we simplify our equation using the distributive property:
y = 2-4(x-1) = 2-4(x) -4(-1) = 2-4x--4 = 2-4x + 4 = -4x + 6
Lines that are parallel have the same slope; this means we want a line through (1, -3) with a slope of -4. Using point-slope form,
#2) We must first write our second equation in slope-intercept form by isolating the y term:
2x+y=7
Subtract 2x from each side"
2x+y-2x = 7-2x
y = -2x+7
This means the slope is -2; the slopes are the same, so the lines are parallel.
#3) The slope of our given equation is 2. To be perpendicular, the second line must have a slope that is a negative reciprocal (flipped and opposite signs); this makes it -1/2. Using point-slope form,
#4) The slope of Main Street on the diagram, found by using rise/run, is 2. This means the bike path will also have a slope of 2 in order to be parallel. The park entrance is at (0, 4). This makes the equation y = 2x+4.
#5) Two lines with the same slope are always parallel. They are never perpendicular.
Answer:
∠ ABC = 53.5°
Step-by-step explanation:
The inscribed angle ABC is half the measure of the intercepted arc AC
∠ ABC = × 107° = 53.5°
<u>Answer:</u>
a) 3.675 m
b) 3.67m
<u>Explanation:</u>
We are given acceleration due to gravity on earth =
And on planet given =
A) <u>Since the maximum</u><u> jump height</u><u> is given by the formula </u>
Where H = max jump height,
v0 = velocity of jump,
Ø = angle of jump and
g = acceleration due to gravity
Considering velocity and angle in both cases
Where H1 = jump height on given planet,
H2 = jump height on earth = 0.75m (given)
g1 = 2.0 and
g2 = 9.8
Substituting these values we get H1 = 3.675m which is the required answer
B)<u> Formula to </u><u>find height</u><u> of ball thrown is given by </u>
which is due to projectile motion of ball
Now h = max height,
v0 = initial velocity = 0,
t = time of motion,
a = acceleration = g = acceleration due to gravity
Considering t = same on both places we can write
where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations
substituting h2 = 18m, g1 = 2.0 and g2 = 9.8
We get h1 = 3.67m which is the required height