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2 years ago

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Approximately 7.5 × 10^5 gallons of water flow over a waterfall each second. There are 8.6 × 10^4 seconds in one day. Approximat

ely how many gallons of water with flow over that waterfall in one day?

1 answer:

tino4ka555 [31]2 years ago

6 0

Answer:

$6.45 \times {10}^{10}$

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Please help me find x<br> Need help fast!!

Bond [772]

Answer:

x = 100°

Step-by-step explanation:

p and q are parallel lines. Construct line 'l' parallel to p and q

a = 30° {p// l when parallel lines are intersected by transversal alternate interior angles are congruent}

c +110 = 180 {linear pair}

c = 180 - 110

c = 70°

b= c {q // l when parallel lines are intersected by transversal alternate interior angles are congruent}

b = 70°

x = a + b

x = 30° + 70°

x = 100°

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3 years ago

How to differentiate y=x^n using the first principle. In this question, I cannot use the rule of differentiation. I have to do t

Zarrin [17]

By first principles, the derivative is

$\displaystyle\lim_{h\to0}\frac{(x+h)^n-x^n}h$

Use the binomial theorem to expand the numerator:

$(x+h)^n=\displaystyle\sum_{i=0}^n\binom nix^{n-i}h^i=\binom n0x^n+\binom n1x^{n-1}h+\cdots+\binom nnh^n$

$(x+h)^n=x^n+nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n$

where

$\dbinom nk=\dfrac{n!}{k!(n-k)!}$

The first term is eliminated, and the limit is

$\displaystyle\lim_{h\to0}\frac{nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n}h$

A power of $h$ in every term of the numerator cancels with $h$ in the denominator:

$\displaystyle\lim_{h\to0}\left(nx^{n-1}+\dfrac{n(n-1)}2x^{n-2}h+\cdots+nxh^{n-2}+h^{n-1}\right)$

Finally, each term containing $h$ approaches 0 as $h\to0$ , and the derivative is

$y=x^n\implies y'=nx^{n-1}$

4 0

3 years ago

How much money should a student place in a time deposit in a bank that pays 1.1% compounded annually so that he will have Php 20

STatiana [176]

Let us say that:<span>
P = present value
F = future value
i = interest rate
n = period

P = F / [ (1 + i ) ^n ]
P = 200000 / [ (1 + 0.011) ^6 ]
P = 187293.65

<span>Therefore the student must put up Php 187,293.65</span></span>

5 0

3 years ago

A 120-inch strip of metal 12 inches wide is to be made into a small open trough by bending up two sides on the long side, at rig

mezya [45]

Answer:

x= 3 inch should be turned up on each side

Step-by-step explanation:

Let the height of trough be x.

Width of trough be 12 - 2x.

and length of trough = 120 inch

Volume of trough, V = L×W×H = 120 × (12-2x) × x = 120x(12 - 2x)

For maximum volume, we find V' = 0

i.e 1440 -480x = 0

or x = $\frac{1440}{480}$

or x= 3

Hence x= 3 inch should be turned up on each side

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3 years ago

How many liters of a 90% acid solution must be added to 6 liters of a 15% acid solution to obtain a 40% acid solution?

tatiyna

3 liters must be added

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3 years ago