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3 years ago

5

The answers x^1 or x^0 How??

1 answer:

pickupchik [31]3 years ago

6 0

Remember a^(m) x a^(n) = a(m+n) & √a = a^(1/2)

===> x^(1/2 -3/2)+x(1/2 -1/2) ====> x^(-1)+x^(0) (any number Exp 0 =1)
X^(-1) +1 ===> 1/x +1 or (1+x)/x

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5+5 you get 39 points

Nina [5.8K]

Answer:

10

Step-by-step explanation:

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3 years ago

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N is an integer with -5 < 2n <6<br> write down all the values of n?

Alika [10]

Answer:

The values of n are less than 3 and more than (-5/2)

Step-by-step explanation:

The given inequality is :

-5 < 2n <6

We need to write the values of n.

Dividing both sides of the inequality by 2. So,

$\dfrac{ -5}{2} < \dfrac{2n}{2}$

So, the values of n are less than 3 and more than (-5/2).

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3 years ago

What is the length of the hypotenuse for a right triangle with legs of length 9 and 40?

Step2247 [10]

Answer:

$c = 41$

Step-by-step explanation:

${c}^{2} = \sqrt{ {a}^{2} + {b}^{2} }$

$c = \sqrt{ {a}^{2} + {b}^{2} }$

$c = \sqrt{ {9}^{2} + {40}^{2} }$

$c = \sqrt{81 + 1600}$

$c = \sqrt{1681}$

$c = 41$

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2 years ago

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A pair of designer sunglasses was marked up 10% and is now selling for $165.00. What was the original price of the sunglasses?

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X+(x*0.1)=165
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3 years ago

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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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3 years ago

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