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3 years ago

Please help. I know it’s easy but I honestly forgot this type of stuff!

Mathematics

2 answers:

solniwko [45]3 years ago

6 0

Answer:

Step-by-step explanation:

I'm pretty sure it is 750 because the area of the pool times the depth is 150*5 which is 750. If it is wrong then sorry

natita [175]3 years ago

5 0

I am sure it’s 750

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Romashka-Z-Leto [24]

f(x)=√(x+9)

f(x+h) =√(x+h+9)

$f'(x) = \lim_{h \to \\0} \frac{\sqrt{x+9+h}-\sqrt{x+9}}{h} = \\ \\ = \lim_{h \to \\0} \frac{(\sqrt{x+9+h}-\sqrt{x+9})*(\sqrt{x+9+h}+\sqrt{x+9})}{h(\sqrt{x+9+h}+\sqrt{x+9})} = \\ \\= \lim_{h \to \\0} \frac{(\sqrt{x+9+h})^{2}-(\sqrt{x+9})^{2}}{h(\sqrt{x+9+h}+\sqrt{x+9})}= \\ \\= \lim_{n \to \\0} \frac{x+9+h-x-9}{h(\sqrt{x+9+h}+\sqrt{x+9})}=\lim_{n \to \\0} \frac{h}{h(\sqrt{x+9+h}+\sqrt{x+9})}=\lim_{n \to \\0} \frac{1}{(\sqrt{x+9+h}+\sqrt{x+9})} = \\ \\= \frac{1}{(\sqrt{x+9+0}+\sqrt{x+9})} =$

$=\frac{1}{(\sqrt{x+9}+\sqrt{x+9})} = \frac{1}{2\sqrt{x+9}}$

$f'(x)= \frac{1}{2\sqrt{x+9}} \\ \\ f'(a)=\frac{1}{2\sqrt{a+9}} \\ \\a=7 \\ \\f'(7)= \frac{1}{2\sqrt{7+9}} = \frac{1}{2*4} =\frac{1}{8} \\ \\f'(7)= \frac{1}{8}$

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Borrowing maybe is right

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