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2 years ago

Evaluate the limits, if they exist. If not, write DNE.

Mathematics

2 answers:

GrogVix [38]2 years ago

6 0

2
DNE
-2 not sure about the answer

Blizzard [7]2 years ago

4 0

A) 2
B) DNE
C) -2

I hope this answers your question but I’m not so sure of my answer!

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Simplify i12. 1 −1 −i i

Likurg_2 [28]

Answer:

Step-by-step explanation:

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2 years ago

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Find the maximum value or minimum value for the function f(x) = 0.15(x + 1)² - 3.

il63 [147K]

Answer:

The minimum value for $f(x) = 0.15(x + 1)^2 - 3$ is $-3$ .

Step-by-step explanation:

Given function is $f(x) = 0.15(x + 1)^2 - 3$

We need to find the maximum value or the minimum value for the function.

Now, differentiate $f(x) = 0.15(x + 1)^2 - 3$ w.r.t $x$ .

$f'(x) =\frac{d}{dx}(0.15(x + 1)^2 - 3)\\f'(x)=\frac{d}{dx}(0.15(x+1)^2-\frac{d}{dx}(3)\\$

$f'(x)=2\times 0.15(x+1)\frac{d}{dx}(x+1)-0\\f'(x)=0.3(x+1)(1)\\f'(x)=0.3(x+1)$

Now, we will equate $f'(x)=0$ to find critical point.

$0.3(x+1)=0\\x=-1$

Plug this critical point in to the function $f(x) = 0.15(x + 1)^2 - 3$ we get,

$f(-1) = 0.15(-1 + 1)^2 - 3\\f(-1)=-3$

Also, $f''(x)=0.3$ which is positive, We have minimum value.

So, the minimum value for $f(x) = 0.15(x + 1)^2 - 3$ is $-3$ .

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3 years ago

Take away 6 from 5 times g ~ in albebraic expression.

Brrunno [24]

5g-6 when you multiply a verable with a number you simplily put the letter beside the number and that lets us know to multiply:)

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2 years ago

If the graph of the linear equation ax-8y=15 has an x-intercept at (3,0), then what is the value of a?

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Ax-8y=15
a(3)-8(0)=15
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a=5

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3 years ago

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Answer:

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Step-by-step explanation:

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