Given x > 0, simplify V25x6 completely.

Mathematics

1 answer:

kirill [66]3 years ago

8 0

Answer:

لو تكرمتم ابحثو لنا عن الاجابة

Step-by-step explanation:

بليز يا شباب

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-3/8a + 3.2b + (-6b) -1/4a I am not sure about this question please help!

expeople1 [14]

Answer:

-5/8a - 2.8b

Step-by-step explanation:

-3/8a + 3.2b + (-6b) - 1/4a

=> -3/8a + 3.2b + (-6b) - 2/8a

=> -5/8a + 3.2b - 6b

=> -5/8a - 2.8b

Therefore, -5/8a - 2.8b is the solution to your problem.

Hoped this helped.

5 0

3 years ago

What is the answer to 2 5/3-2 3/2=

jek_recluse [69]

First, we can convert both of them to improper fractions.

We do that by multiplying the denominator to the whole number, adding it to the numerator, and keeping the denominator.

2 5/3 - 2 3/2

So we have:

11/3 - 7/2

Convert both of them to denominators of 6:

22/6 - 21/6

Subtract the numerators and keep the denominators:

1/6

3 0

3 years ago

Read 2 more answers

At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

MAXImum [283]

Answer-

At $x= \frac{1}{2304e^4-16e^2}$ the curve has maximum curvature.

Solution-

The formula for curvature =

$K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}$

Here,

$y=4e^{x}$

Then,

${y}' = 4e^{x} \ and \ {y}''=4e^{x}$

Putting the values,

$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$