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DiKsa [7]
3 years ago
7

Which equation of c makes this equation true. 12-9+c =12 Pls only direct answers

Mathematics
2 answers:
alexandr1967 [171]3 years ago
8 0

Answer:

c or 9

Step-by-step explanation:

Naya [18.7K]3 years ago
3 0

Answer:

C: 9

Step-by-step explanation:

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A boat travels 33 miles downstream in 4 hours. The return trip takes the boat 7 hours. Find the speed of the boat in still water
Paha777 [63]

<u>Answer:</u>

Speed of the boat in still water = 6.125 miles/hour

<u>Step-by-step explanation:</u>

We are given that a boat travels 33 miles downstream in 4 hours and the return trip takes the boat 7 hours.

We are to find the speed of the boat in the still water.

Assuming S_b to be the speed of the boat in still water and S_w to be the speed of the water.

The speeds of the boat add up when the boat and water travel in the same direction.

Speed = \frac{distance}{time}

S_b+S_w=\frac{d}{t_1}=\frac{33 miles}{4 hours}

And the speed of the water is subtracted from the speed of the boat when the boat is moving upstream.

S_b-S_w=\frac{d}{t_2}=\frac{33 miles}{7 hours}

Adding the two equations to get:

   S_b+S_w=\frac{d}{t_1}

+  S_b-S_w=\frac{d}{t_2}

___________________________

2S_b=\frac{d}{t_1} +\frac{d}{t_2}

Solving this equation for S_b and substituting the given values for d,t_1, t_2:

S_b=\frac{(t_1+t_2)d}{2t_1t_2}

S_b=\frac{(4 hour + 7hour)33 mi}{2(4hour)(7hour)}

S_b=\frac{(11 hour)(33mi)}{56hour^2}

S_b=6.125 mi/hr

Therefore, the speed of the boat in still water is 6.125 miles/hour.

6 0
3 years ago
Read 2 more answers
Use the disk method or the shell method to find the volumes of the solids generated by revolving the region bounded by the graph
diamong [38]

Answer:

a) 8π

b) 8/3 π

c) 32/5 π

d) 176/15 π

Step-by-step explanation:

Given lines :  y = √x, y = 2, x = 0.

<u>a) The x-axis </u>

using the shell method

y = √x = , x = y^2

h = y^2 , p = y

vol = ( 2π ) \int\limits^2_0 {ph} \, dy

     = ( 2\pi ) \int\limits^2_0 {y.y^2} \, dy  

∴ Vol = 8π

<u>b) The line y = 2  ( using the shell method )</u>

p = 2 - y

h = y^2

vol = ( 2π ) \int\limits^2_0 {ph} \, dy

     = ( 2\pi ) \int\limits^2_0 {(2-y).y^2} \, dy

     = ( 2π ) * [ 2/3 * y^3  - y^4 / 4 ] ²₀

∴ Vol  = 8/3 π

<u>c) The y-axis  ( using shell method )</u>

h = 2-y  = h = 2 - √x

p = x

vol = (2\pi ) \int\limits^4_0 {ph} \, dx

     = (2\pi ) \int\limits^4_0 {x(2-\sqrt{x}  ) } \, dx

     = ( 2π ) [x^2 - 2/5*x^5/2 ]⁴₀

vol = ( 2π ) ( 16/5 ) = 32/5 π

<u>d) The line x = -1    (using shell method )</u>

p = 1 + x

h = 2√x

vol = (2\pi ) \int\limits^4_0 {ph} \, dx

Hence   vol = 176/15 π

attached below is the graphical representation of P and h

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70,000 when a zero is in between you gonna count it
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