Answer:
Step-by-step explanation:
The equation of a straight line can be represented in the slope intercept form as
y = mx + c
Where
m = slope = (change in the value of y in the y axis) / (change in the value of x in the x axis)
The equation of the given line is
x + 2y = 4
2y = - x + 4
y = -x/2 + 4/2
y = - x/2 + 2
Comparing with the slope intercept form, slope = - 1/2
If two lines are perpendicular, it means that the slope of one line is the negative reciprocal of the slope of the given line.
Therefore, the slope of the line passing through (- 2, 1) is 2
To determine the intercept, we would substitute m = 2, x = - 2 and y = 1 into y = mx + c. It becomes
1 = 2 × - 2 + c = - 4 + c
c = 1 + 4 = 5
The equation becomes
y = 2x + 5
Answer:
-3v
Step-by-step explanation:
-6v is a negative when 3v is positive
when you add 3v to -6v it equals -3v since 6 is greater
Answer:
112
Step-by-step explanation:
To find the least common multiple, we break the numbers down into prime factors
7 = 7
14 = 7*2
16 = 2*8 = 2*4*2 = 2*2*2*2
Then we take the largest number of times it appears in all the numbers
2 appears 4 times so we take take four 2's
7 appears 1 so we take one 7
LCM = 2*2*2*2*7
LCM = 112
Answer:
x = 3
y = 15
Step-by-step explanation:
If △XPS ≅△DNF, their corresponding sides would be congruent. This implies that:
XP ≅ DN
PS ≅ NF
XS ≅ DF
Given that:
XP = 4y - 3
DN = 57
NF = 51
XS = 17x + 3
DF = 54
Therefore:
XP = DN
4y - 3 = 57 (Substitution)
Add 3 to both sides
4y = 57 + 3
4y = 60
Divide both sides by 4
y = 60/4
y = 15
Also,
XS = DF
17x + 3 = 54 (substitution)
Subtract 3 from each side
17x = 54 - 3
17x = 51
Divide both sides by 17
x = 51/17
x = 3
it takes 5 seconds to reach a height of 40 ft. and It takes 2.5 seconds to reach maximum height.
Let h(t) represent the height of the ball at time t.
Given that the height is given by:
h(t) = -16t² + 80t + 40
1) For the ball to reach a height 0f 40 ft, h(t) = 40, hence:
40 = -16t² + 80t + 40
16t² - 80t = 0
t(t - 5) = 0
t = 0 or t = 5
Hence it takes 5 seconds to reach a height of 40 ft.
2) The maximum height is at h'(t) = 0,
h'(t) = -32t + 80
-32t + 80 = 0
32t = 80
t = 2.5
It takes 2.5 seconds to reach maximum height.
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