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xenn [34]
2 years ago
12

Please help it would be much appreciated!!

Mathematics
1 answer:
AlekseyPX2 years ago
3 0

Answer:

a) P) 0.25 = 1/4  both

b) P) 0.75 = 3/4  one

c) P) 0.25  = 1/4  none

Step-by-step explanation:

4 routes to D + H

6 routes to H + S

= D  H   S

     4  | 6

    - 2 | -3

=    2  |  3  = route 1 = 2/4  and route 2 = 3/6

= 1/2 route 1 and 1/2 route 2

Answer a ) = 1/2 x 1/2 = 0.25 = 1/4  = 0.25 probability

Answer b)  = 1- (1/2 x 1/2) = 1- 0.25 = 3/4 = 0.75 probability

Answer c)  = 1/2 x 1/2 = 0.25 = 1/4 = 0.25 probability  

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What is being constructed based on the markings in the following diagram?
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  perpendicular line through a point on a line

Step-by-step explanation:

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2 years ago
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Find the volume of the solid by rotating the region bounded by y=x^3, y=8, and x=0 about the y axis.
dimulka [17.4K]

Answer:

Solution : Volume = 96/5π

Step-by-step explanation:

If we slice at an arbitrary height y, we get a circular disk with radius x, where x = y^(1/3). So the area of a cross section through y should be:

A(y) = πx^2 = π(y^(1/3))^2 = πy^(2/3)

And now since the solid lies between y = 0, and y = 8, it's volume should be:

V =  ∫⁸₀  A(y)dy (in other words ∫ A(y)dy on the interval [0 to 8])

=> π ∫⁸₀ y^(2/3)dy

=> π[3/5 * y^(5/3)]⁸₀

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=> 96/5π ✓

7 0
3 years ago
Solve irrational equation pls
rusak2 [61]
\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)



\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\

4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\

There's one more condition I forgot about
-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\

Finally
x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}
3 0
2 years ago
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