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Find the area of a trapezium whose parallel sides are respectively 14cm and 4cm and distance between the parallel sides is 5cm.
1 answer:
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Answer:
Value of the independent variable = 25
Step-by-step explanation:
y-intercept form of a straight line is represented by the equation,
y = mx + b
here y = dependent variable
m = slope
b = y intercept or vertical intercept
x = independent variable
In our question,
y-intercept 'b' = 150
slope of the line 'm' = 4
dependent variable y = 250
We will plug in these values in the equation of the line to get the value of the independent variable.
250 = 4x + 150
250 - 150 = 4x
x =
x = 25
Therefore, value of the independent variable is 25.
Your answer to this is :
<h2>C. Stretching.</h2><h2>→ <u>Explanation:-</u></h2>
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Because the other one is congruent together.
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<em>Jaceysan ~</em>
Answer:
f(x) = 3 cos2(x) − 6 sin(x), 0 ≤ x ≤ 2π
We plot the function using a graphing calculator (See attached image below)
a) 1. Find the interval on which f is increasing.
f(x) increases for
x ∈ [π/2 , 7π/6] ∪ [3π/2, 11π/6]
2. Find the interval on which f is decreasing
f(x) decreases for
x ∈ [0 , π/2) ∪ (7π/6, 3π/2) ∪ (11π/6, 2π]
(b) Find the local minimum and maximum values of f
Local minimum. f = -9
Local maximum. f = 4.5
(c) 1. Find the inflection points
Please see second image attached
Points,
(x,y) = (2.50673, -2.66882)
(x,y) = (4.14456, 3.79382)
(x,y) = (5.28022, 3.79382)
2. Find the interval on which f is concave up.
x ∈ [0 , 2.50673] ∪ [4.14456, 5.28022]
3. Find the interval on which f is concave down.
x ∈ (2.50673, 4.14456) ∪ (5.28022, 2π]
