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2 years ago

14

Use a table of values to estimate the value of the limit. Confirm your result graphically by graphing the function with a graphi

ng device. . If the limit does not exist enter DNE. . . \lim_{x\to 0} \frac{\sqrt{x+25}- 5}{x}

1 answer:

il63 [147K]2 years ago

4 0

Hello,

$\lim_{x \to 0}\dfrac{ \sqrt{x+25}-5}{x}\\\\$

$=\lim_{x \to 0}\dfrac{(\sqrt{x+25}-5)*(\sqrt{x+25}+5)}{x*(\sqrt{x+25}+5)}\\\\$

$=\lim_{x \to 0}\dfrac{x+25-25)}{x*(\sqrt{x+25}+5)}\\\\$

$=\dfrac{1}{10}$

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Determinar el área de un hexágono que esta inscrito en una circunferencia de 4 unidades de radio

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Answer:

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2 years ago

Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st

Naily [24]

Answer:

$\boxed{2^{\frac{802}{27}} \cdot 3^9}$

Step-by-step explanation:

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$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

$\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }$

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

$\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3$

Also,

$\boxed{a^{0} = 1, a\neq 0 }$

$\left(\dfrac{1}{9} \cdot \dfrac{1}{1024} \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

Note

$\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

Note

$\boxed{\dfrac{1}{\dfrac{1}{a} } = a}$

$9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)$

Once

$9216=2^{10}\cdot 3^2 \implies 9216^3=2^{30}\cdot 3^6$

$\boxed{(a \cdot b)^n=a^n \cdot b^n}$

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27}$

We have

$\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)$

Also, once

$\boxed{\dfrac{c^a}{c^b}=c^{a-b}}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27$

As

$30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27} =\dfrac{802}{27}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3$

$2^{\frac{802}{27}} \cdot 3^9$

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2 years ago

Complete the table.

Sauron [17]

Answer:

40

60

Step-by-step explanation:

20÷2=10

10×4=40

10×6=60

3 0

2 years ago

Read 2 more answers

you should have $16 and a coupon for a $5 discount at a local supermarket. a bottle of olive oil costs$7. how many bottles of ol

natta225 [31]

Each bottle costs seven dollars. Two of them would cost $14 and three would cost $21. Buying three bottles, the amount is $21 which exceeds $16. Therefore, a $5 discount will be given. Subtracting $5 from the $21 will give $16. You have just enough amount to buy THREE bottles of olive oil.

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2 years ago

52,908 round to the nearest hundred.

IrinaVladis [17]

The hundredth is the 3rd number after the decimal place (so the 9). Look at the number before it and if it is 5 or above then round it to the next hundredth. In this case it is only a zero so we don't change the 9 just substitute it 0's. Therefore the answer is 52,900.

If you need further help let me know, hope this helped you.

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2 years ago