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tensa zangetsu [6.8K]
3 years ago
10

Please help with this problem

Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
4 0

Answer:

that looks hard

Step-by-step explanation:

that's what she said

lol

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Lim x-> 0 sqrt x+4 -2 / x
Wittaler [7]

Greetings from Brasil...

Here we have an indeterminacy 0/0

We can change variable or rationalize.......

Let's rationalize

{[√(X + 4) - 2]/X} · {[√(X + 4) + 2]/[√(X + 4) + 2]} = 1/[√(X + 4) + 2]

So, the limit will be 1/4

3 0
1 year ago
at a restaurant 6 hot dogs cost $8.28 and 7 hamburgers cost $9.24. which food has the lower unit price?
SIZIF [17.4K]

Answer:

Step-by-step explanation: Hamburgers

3 0
3 years ago
Read 2 more answers
Va rog e urgent am nevoie de raspuns
Lady_Fox [76]

Answer: 4

<u>Explanation:</u>

f(x) = 2x - 1

f(√2) = 2√2 - 1

f(1) = 2(1) - 1

     =  2 - 1

     =     1

f(√3)  = 2√3 - 1

*******************************************************

\frac{f(\sqrt{2})-f(1)}{\sqrt{2}-1} +\frac{f(\sqrt{3})-f(\sqrt{2})}{\sqrt{3}-\sqrt{2}}

= \frac{(2\sqrt{2}-1)-1}{\sqrt{2}-1} +\frac{(2\sqrt{3}-1)-(2\sqrt{2}-1)}{\sqrt{3}-\sqrt{2}}

= \frac{2\sqrt{2}-2}{\sqrt{2}-1} +\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}

= \frac{2\sqrt{2}-2}{\sqrt{2}-1}(\frac{\sqrt{2}+1}{\sqrt{2}+1})+\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}})

= \frac{4+2\sqrt{2}-2\sqrt{2}-2}{2 - 1} + \frac{6 +2\sqrt{6}-2\sqrt{6}-4}{3-2}

= \frac{2}{1} +\frac{2}{1}

= 4

3 0
3 years ago
What is the answer to this problem? <br><br> 2 1/3 + 3 1/3 =?
Naily [24]

Answer:

Convert the mixed numbers to improper fractions, then find the LCD and combine.

Exact Form:

173

Decimal Form:

5.¯6

Mixed Number Form:

523

Step-by-step explanation:

3 0
3 years ago
Describe how to decide if 94 is a prime number or composite number
algol [13]
94 is a composite number.

If 94 were a prime number, 94 would only have 2 factors, which are 1 and itself. Since 94 has another factor other than 1 and itself (which is 2), 94 is a composite number.

Hope this helps~
6 0
3 years ago
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