In polar coordinate, <em>R</em> is the set of points
{(<em>r</em>, <em>θ</em>) | 1 < <em>r</em> < 5 and 0 < <em>θ</em> < <em>π</em>/2}
So the integral is
(where <em>s</em> = <em>r</em> ²)
Answer:3
Step-by-step explanation:
Put it onto number line and it is 3
Answer:
9 dolloars an hour man thats great pay
Step-by-step explanation:
Factor the following:
x^2 - 5 x - 24
Hint: | Factor x^2 - 5 x - 24 by finding two numbers whose product is -24 and whose sum is -5.
The factors of -24 that sum to -5 are 3 and -8. So, x^2 - 5 x - 24 = (x + 3) (x - 8):
Answer: (x + 3) (x - 8)
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Factor the following:
49 x^2 - 64
Hint: | Express 49 x^2 - 64 as a difference of squares.
49 x^2 - 64 = (7 x)^2 - 8^2:
(7 x)^2 - 8^2
Hint: | Factor the difference of two squares.
Factor the difference of two squares. (7 x)^2 - 8^2 = (7 x - 8) (7 x + 8):
Answer: (7 x - 8) (7 x + 8)
<span>-24 + 15 x + x^2
</span>
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Factor the following:
x^2 - x - 110
Hint: | Factor x^2 - x - 110 by finding two numbers whose product is -110 and whose sum is -1.
The factors of -110 that sum to -1 are 10 and -11. So, x^2 - x - 110 = (x + 10) (x - 11):
Answer: (x + 10) (x - 11)___________________________________________________________
Factor the following:
49 x^2 - 64
Hint: | Express 49 x^2 - 64 as a difference of squares.
49 x^2 - 64 = (7 x)^2 - 8^2:
(7 x)^2 - 8^2
Hint: | Factor the difference of two squares.
Factor the difference of two squares. (7 x)^2 - 8^2 = (7 x - 8) (7 x + 8):
Answer: (7 x - 8) (7 x + 8)
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Factor the following:
25 x^2 + 40 x - 10
Hint: | Factor out the greatest common divisor of the coefficients of 25 x^2 + 40 x - 10.
Factor 5 out of 25 x^2 + 40 x - 10:
Answer: 5 (5 x^2 + 8 x - 2)
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Factor the following:
9 x^2 + 3 x^2 - 6 x
Hint: | Add like terms in 9 x^2 + 3 x^2 - 6 x.
3 x^2 + 9 x^2 = 12 x^2:
12 x^2 - 6 x
Hint: | Factor common terms out of 12 x^2 - 6 x.
Factor 6 x out of 12 x^2 - 6 x:
Answer: 6 x (2 x - 1)
ΔQTR and ΔSTP are similar triangles. The corresponding sides of similar triangles are the same length.
TR = PT
TS = QT
Input the algebraic expression to make an equation system
PT = TR
y = 2x - 1 <em>(first equation)</em>
TS = QT
6x + 13 = 5y
6x - 5y = -13 <span><em>(second equation)</em>
</span>Using subtitution method, subtitute y with 2x - 1 from first equation into the second equation, to find the value of x
6x - 5y = -13
6x - 5(2x - 1) = 13
6x - 10x + 5 = 13
-4x + 5 = 13
-4x = 13 - 5
-4x = 8
x = 8/-4
x = -2
Find y with subtituting x with -2 in the first equation
y = 2x - 1
y = 2(-2) - 1
y = -4 - 1
y = -5
Answer:
x = -2
y = -5
