Answer:
- The square root and quadratic function share a y-intercept.
- The range of the square root and absolute value function are the same.
Step-by-step explanation:
Y-intercepts are the same when the curves meet the y-axis at the same point. That is true of the root and quadratic functions.
X-intercepts are the same when the curves meet the x-axis at the same point. None of these functions share an x-intercept.
The ranges of the functions are the same when they have the same vertical extent. The range of the quadratic is different from the range of the other two functions.
The absolute value and root functions have the same minimum (lower end of their range). That is the same as the maximum of the quadratic function.
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The statements that match the graphs are ...
- The square root and quadratic function share a y-intercept.
- The range of the square root and absolute value function are the same.
Answer
15 + (-3s) or 15 - 3s
Step-by-step explanation:
-3 times -5 is 15
-3 times s is -3s
Answer:
You may or may not need to include the units.
A = 18x - 18
P = 6x + 6
Graph is attached below. (2, 18)
Step-by-step explanation:
Substitute the information we need, "l" and "w", into the formulas.
l is for length, 6cm.
w is for width, (3x - 3)cm.
Use the formula for area of a rectangle.
A = lw
A = (6)(3x-3)cm²
A = (18x - 18)cm² or 18x - 18
Use the formula for perimeter of a rectangle.
P = 2(l + w)
P = 2(6 + (3x - 3))cm
P = 2(3x + 3)cm
P = (6x + 6)cm or 6x + 6
Linear equations are written in the form y = mx + b, so we do not need to factor or further simplify the formulas.
To graph, first turn the "m" value into a fraction form.
8 -> 8/1
6 -> 6/1
You need two points to graph each line.
For each equation, the first point is on the y-axis at the "b" value. Then use the "m" in the equation to count the number of units up (numerator) and to the right (denominator).
The solution is (2,18)
1. cot(x)sec⁴(x) = cot(x) + 2tan(x) + tan(3x)
cot(x)sec⁴(x) cot(x)sec⁴(x)
0 = cos⁴(x) + 2cos⁴(x)tan²(x) - cos⁴(x)tan⁴(x)
0 = cos⁴(x)[1] + cos⁴(x)[2tan²(x)] + cos⁴(x)[tan⁴(x)]
0 = cos⁴(x)[1 + 2tan²(x) + tan⁴(x)]
0 = cos⁴(x)[1 + tan²(x) + tan²(x) + tan⁴(4)]
0 = cos⁴(x)[1(1) + 1(tan²(x)) + tan²(x)(1) + tan²(x)(tan²(x)]
0 = cos⁴(x)[1(1 + tan²(x)) + tan²(x)(1 + tan²(x))]
0 = cos⁴(x)(1 + tan²(x))(1 + tan²(x))
0 = cos⁴(x)(1 + tan²(x))²
0 = cos⁴(x) or 0 = (1 + tan²(x))²
⁴√0 = ⁴√cos⁴(x) or √0 = (√1 + tan²(x))²
0 = cos(x) or 0 = 1 + tan²(x)
cos⁻¹(0) = cos⁻¹(cos(x)) or -1 = tan²(x)
90 = x or √-1 = √tan²(x)
i = tan(x)
(No Solution)
2. sin(x)[tan(x)cos(x) - cot(x)cos(x)] = 1 - 2cos²(x)
sin(x)[sin(x) - cos(x)cot(x)] = 1 - cos²(x) - cos²(x)
sin(x)[sin(x)] - sin(x)[cos(x)cot(x)] = sin²(x) - cos²(x)
sin²(x) - cos²(x) = sin²(x) - cos²(x)
+ cos²(x) + cos²(x)
sin²(x) = sin²(x)
- sin²(x) - sin²(x)
0 = 0
3. 1 + sec²(x)sin²(x) = sec²(x)
sec²(x) sec²(x)
cos²(x) + sin²(x) = 1
cos²(x) = 1 - sin²(x)
√cos²(x) = √(1 - sin²(x))
cos(x) = √(1 - sin²(x))
cos⁻¹(cos(x)) = cos⁻¹(√1 - sin²(x))
x = 0
4. -tan²(x) + sec²(x) = 1
-1 -1
tan²(x) - sec²(x) = -1
tan²(x) = -1 + sec²
√tan²(x) = √(-1 + sec²(x))
tan(x) = √(-1 + sec²(x))
tan⁻¹(tan(x)) = tan⁻¹(√(-1 + sec²(x))
x = 0
Quarterly or monthly because semi annually is to soon do I think monthly
