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2 years ago

Please help! A, B, C, D, or E?

Mathematics

1 answer:

Leviafan [203]2 years ago

4 0

I) This is False, because as you approach -3 from the right, the denominator is positive making g(x) negative, thus the limit is negative infinity.

$\lim_{x \to -3^+} g(x) = - \infty$

II) This is True, as x goes to infinity, cos(x) is bounded between -1 and 1 and can be treated as a constant. Therefore the limit is 1.

$\lim_{x \to \infty} \frac{x+ c}{x+3} = \frac{x}{x} = 1$

III) This is False, the intermediate value theorem only applies on intervals where function is continuous. g(x) is not defined at x=-3, which is in the given interval [-4,0]. Therefore g(x) is not continuous and intermediate value theorem does not apply.

Final Answer:

D) only statement II is True

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2 years ago

Read 2 more answers

How to Factor 4p^2+4p-5

Anarel [89]

4p^2 + 4p - 5

As the expression cannot be factored in it's current form, we need to check if splitting the middle term is possible.

List factors of -20.

-5 * 4

5 * -4

-10 * 2

10 * -2

-20 * 1

20 * -1

None of the factors added together result in positive 4.

Because splitting the middle term is not possible, and we cannot simplify this expression any further:

This expression cannot be fully factored.

7 0

2 years ago

In the diagram below segment DEF is shown(since all three letters are shown, they are collinear). If DE=3x+2, EF= x+5, and DF= 2

amm1812

The value of $x$ is 4.

By Geometry, we know that line segment $DF = DE + EF$ , which is equivalent to:

$DF = 3\cdot x + 2 + x + 5$

$DF = 4\cdot x + 7$ (1)

Now we solve algebraically the resulting expression.

If we know that $DF = 23$ , then we solve the equation for $x$ :

$4\cdot x + 7 = 23$

$4\cdot x = 16$

$x = 4$

The value of $x$ is 4.

We kindly invite to see this question on line segments: brainly.com/question/23297288

6 0

2 years ago

Show 2 different solutions to the task.

laila [671]

Answer with Step-by-step explanation:

1. We are given that an expression $n^2+n$

We have to prove that this expression is always is even for every integer.

There are two cases

1.n is odd integer

2.n is even integer

1.n is an odd positive integer

n square is also odd integer and n is odd .The sum of two odd integers is always even.

When is negative odd integer then n square is positive odd integer and n is negative odd integer.We know that difference of two odd integers is always even integer.Therefore, given expression is always even .

2.When n is even positive integer

Then n square is always positive even integer and n is positive integer .The sum of two even integers is always even.Hence, given expression is always even when n is even positive integer.

When n is negative even integer

n square is always positive even integer and n is even negative integer .The difference of two even integers is always even integer.

Hence, the given expression is always even for every integer.

2.By mathematical induction

Suppose n=1 then n= substituting in the given expression

1+1=2 =Even integer

Hence, it is true for n=1

Suppose it is true for n=k

then $k^2+k$ is even integer

We shall prove that it is true for n=k+1

$(k+1)^1+k+1$

= $k^1+2k+1+k+1$

= $k^2+k+2k+2$

=Even +2(k+1)[/tex] because $k^2+k$ is even

=Sum is even because sum even numbers is also even

Hence, the given expression is always even for every integer n.

3 0

2 years ago