Two lines are perpendicular if and only if the product of their slopes is - 1.
So, you just need to find the slope of each line and find out the product of their slopes.
I will do one example for you.
L1: y = 3x + 5
L2: y = - 3x + 14
L3: y = -x/3 + 14
The slope of a line is the coefficient of the x.
So the slopes are:
L1: slope 3
L2: slope -3
L3: slope -1/3
So now multiply the slopes of each pair of lines:
L1 and L2: 3 * (-3) = - 9 => No, they are not perpendicular
L2 and L3: (-3) * (-1/3) = 1 => No, they are not perpendicular
L1 and L3: (3) * (-1/3) = -1 => Yes, they are penpendicular.
The ball will be at height 19 feet first at 0.7 sec and then at 1.42 sec.
<u>Explanation:</u>
We need to find the time at which the ball will be at height 19 feet.
Equation:
h = 3 + 34t - 16t²
19 = 3 + 34t - 16t²
16 = -16t² + 34t
-16t² + 34t - 16 = 0
On solving the equation, we get
t1 = 0.7 s and t2 = 1.42s
Therefore, the ball will be at height 19 feet first at 0.7 sec and then at 1.42 sec.
The expression as a power is x¹⁰.
Given, the expression is xx⁴x⁴x
In exponents we have the power rule while multiplying.
the rule is that when bases are same, powers are added.
Exponent multiplication is the process of multiplying two terms with exponents. Depending on the base and the power, specific rules apply when multiplying exponents.
⇒ xx⁴x⁴x
base is same, power should be added.
⇒ x¹⁺⁴⁺⁴⁺¹
= x¹⁰
Hence the expression as a power is x¹⁰.
Learn more about Exponents here:
brainly.com/question/11975096
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Answer:
k = -10
Explanation:
f(x) = 4x³ - 3kx² + 44x - 30
The divisor is:
Set divisor to zero:
Insert this into equation.
⇒ 4(2)³ - 3k(2)² + 44(2) - 30 = 210
⇒ 32 - 12k + 88 - 30 = 210
⇒ -12k + 90 = 210
⇒ -12k = 210 - 90
⇒ -12k = 120
⇒ k = 120/-12
⇒ k = -10
Answer:
8
Step-by-step explanation:
To find the mean, add all the numbers together and divide by the amount of numbers there are.
(3 + 3 + 4 + 4 + 8 + 8 + 8 + 10 + 12 + 20)/10 = answer
Simplify. First, solve the parenthesis, then divide.
(80)/10 = answer
8 = answer
8 is the mean of the data set.
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