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likoan [24]
3 years ago
15

F(x) = x2 + 1 g(x) = 5 – x

Mathematics
1 answer:
xz_007 [3.2K]3 years ago
7 0

Answer:

g(x) = 5 – x

Step-by-step explanation:

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Multiply the following numbers. Reduce the lowest terms.
NeX [460]

Answer:

What are the numbers?

Step-by-step explanation:

5 0
2 years ago
Help please!!!!!!!!!
In-s [12.5K]

ANSWER

24


EXPLANATION

For a matrix A of order n×n, the cofactor C_{ij} of element a_{ij} is defined to be


   C_{ij} = (-1)^{i+j} M_{ij}


M_{ij} is the minor of element a_{ij} equal to the determinant of the matrix we get by taking matrix A and deleting row i and column j.


Here, we have


   C_{11} = (-1)^{1+1} M_{11} = M_{11}


M₁₁ is the determinant of the matrix that is matrix A with row 1 and column 1 removed. The bold entries are the row and the column we delete.


   \begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right)  \end{aligned}


Since the determinant of a 2×2 matrix is


   \det\left(  \begin{bmatrix} a & b \\ c& d  \end{bmatrix} \right) = ad-bc


it follows that


   \begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \\ &= (0)(-8) - (-3)(8) \\ &= -(-24) \\ &= 24 \end{aligned}


so C_{11} = M_{11} = 24

4 0
3 years ago
The long jump record, in feet, at a particular school can be modeled by f(x) = 19.6 + 2.5ln(x + 1) where x is the number of year
Sedbober [7]

Answer:

25.8

Step-by-step explanation:

6 0
3 years ago
Calculus piecewise function. ​
Kipish [7]

Part A

The notation \lim_{x \to 2^{+}}f(x) means that we're approaching x = 2 from the right hand side (aka positive side). This is known as a right hand limit.

So we could start at say x = 2.5 and get closer to 2 by getting to x = 2.4 then to x = 2.3 then 2.2, 2.1, 2.01, 2.001, etc

We don't actually arrive at x = 2 itself. We simply move closer and closer.

Since we're on the positive or right hand side of 2, this means we go with the rule involving x > 2

Therefore f(x) = (x/2) + 1

Plug in x = 2 to find that...

f(x) = (x/2) + 1

f(2) = (2/2) + 1

f(2) = 2

This shows \lim_{x \to 2^{+}}f(x) = 2

Then for the left hand limit \lim_{x \to 2^{-}}f(x), we'll involve x < 2 and we go for the first piece. So,

f(x) = 3-x

f(2) = 3-2

f(2) = 1

Therefore, \lim_{x \to 2^{-}}f(x) = 1

===============================================================

Part B

Because \lim_{x \to 2^{+}}f(x) \ne \lim_{x \to 2^{-}}f(x) this means that the limit \lim_{x \to 2}f(x) does not exist.

If you are a visual learner, check out the graph below of the piecewise function. Notice the gap or disconnect at x = 2. This can be thought of as two roads that are disconnected. There's no way for a car to go from one road to the other. Because of this disconnect, the limit doesn't exist at x = 2.

===============================================================

Part C

You'll follow the same type of steps shown in part A.

However, keep in mind that x = 4 is above x = 2, so we'll deal with x > 2 only.

So you'd only involve the second piece f(x) = (x/2) + 1

You should find that f(4) = 3, and that both left and right hand limits equal this value. The left and right hand limits approach the same y value. The limit does exist here. There are no gaps to worry about when x = 4.

===============================================================

Part D

As mentioned earlier, since \lim_{x \to 4^{+}}f(x) = \lim_{x \to 4^{-}}f(x) = 3, this means the limit \lim_{x \to 4}f(x) does exist and it's equal to 3.

As x gets closer and closer to 4, the y values are approaching 3. This applies to both directions.

4 0
1 year ago
Robert threw a rock off a bridge into the river. The distance from the rock to the river is modeled by the equation h=-16ť – 16t
cluponka [151]

Answer:

1.5 seconds

Step-by-step explanation:

h=-16t² – 16t+ 60

When the rock enter the water, the height is 0. ... h=0

0 = -16t² – 16t+ 60 = -4 * (4t² - 4t +15) = -4* (2t - 3) (2t +5)

2t-3=0   or 2t+5 = 0

t = 3/2  or t = -5/2   (time can't be negative)

t = 3/2   or 1.5

8 0
2 years ago
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