Helpp the topic is about limits

Mathematics

1 answer:

erik [133]2 years ago

8 0

Since we want to find the value of k where the limit exists, set both equations equal to each other. Then substitute x = -1 in for each equation to find k.

$kx - 3 = {x}^{2} + k$

1. Set both equations equal.

$- k - 3 = 1 + k$

2. Substitute x = -1.

$- 3 = 1 + 2k$

3. Solve for k by adding k to both sides. Continue the process of solving the equation.

$- 4 = 2k$

$k = - 2$

Thus, k = -2. Check by graphing the function.

You might be interested in

(-7x∧5 + 14 - 2x) + (10x∧4 + 7x + 5x∧5)

Maslowich

$-7x^5 + 5x^5 +10x^4 -2x + 7x + 14= -2x^5 +10x^4 +5x + 14$

Your answer is $-2x^5 +10x^4 +5x + 14$

7 0

3 years ago

as a gift, you fill the calendar with packets of chocolate candy. Each packet has a volume of 2 cubic inches. Find the maximum n

qaws [65]

The picture of the calendar is shown in the attached image.

Now, first we will get the volume of the calendar itself, we can note the calendar has the shape of a triangular prism.
Volume of triangular prism = area of base * depth
The area of base = area of triangle = 1/2 * base * depth
Therefore:
Volume of prism = 1/2 * base * height * depth
where:
base = 4 in
height is he height of the base = 6 in
depth is the depth of the calendar = 8 in
Therefore:
Volume of calendar = 1/2 * 4 * 6 * 8 = 96 in^3

Now, we are given that the volume of each candy is 2 in^3, this means that:
number of candies to fill the calendar = volume of calendar / volume of candy
= 96/2
= 48 candies

Hope this helps :)