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lisabon 2012 [21]
3 years ago
15

Solve this equation, and show work. worth 40 points.

Mathematics
2 answers:
Naily [24]3 years ago
7 0

Answer:

b=-3

Step-by-step explanation:

We have:

25^b\cdot 125^{b+2}=125^b

Notice that 125 is the same as 5³. So:

25^b\cdot ((5)^3)^{b+2}=((5)^3)^b

Also, 25 is the same as 5². So:

((5)^2)^b\cdot ((5)^3)^{b+2}=((5)^3)^b

Power of a power property:

5^{2b}\cdot 5^{3(b+2)}=5^{3b}

Power of a product property:

5^{2b+3(b+2)}=5^{3b}

Since they have the same base, the exponents are equal. So:

2b+3(b+2)=3b

Solve for b. Distribute on the left:

2b+3b+6=3b

Combine like terms on the left:

5b+6=3b

Subtract 5b from both sides:

6=-2b

Divide both sides by -2. So, the value of b is:

b=-3

And we're done!

krek1111 [17]3 years ago
4 0

Answer:

b = - 3

Step-by-step explanation:

Please see the attachment below of the following work.

Thanks! Can I have BRAINLIEST please?

Hope this helps!

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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.

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Anton [14]

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Step-by-step explanation:

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Step-by-step explanation:

1894+x^2

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Adding 213 makes it 1894.

And you end up with the expression 1894+x^2

Hope this helps!

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Step-by-step explanation:

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