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fenix001 [56]
3 years ago
10

Estimate the area of the cottonwood leaf.

Mathematics
1 answer:
elixir [45]3 years ago
4 0
There is no picture to provide an answer to
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The table represents the function f(x). If g(x) = -(x + 1)^2 - 10, which statement is true?
miskamm [114]

Answer:

Answer C is correct.

Step-by-step explanation:

f(x) clearly has a maximum:  y = +10 at x = 0.

Analyzing g(x) = -(x + 1)^2 - 10, we see that the vertex is at (-1, -10), and that the graph opens down.  Thus, -10 is the maximum value; it occurs at x = -1.

Answer A is false.  Both functions have max values.

Answer B is false.  One max is y = 10 and the other is y = -10.

Answer C is correct.  The max value of f(x), which is 10, is greater than the max value of g(x), which is -10.

Answer D is false.  See Answer B, above.

5 0
3 years ago
Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t
Svet_ta [14]

Answer:

[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}

[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}

[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]

Finally, the third vector of basis E is e3(t)=t^{2}

T[e3(t)]=T(t^{2})

in the equation : a2 = 1

T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}

Then

[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]

And that is the third column of [T]EE

Let's write our matrix

[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

X=\left[\begin{array}{c}1&0&0\\\end{array}\right]

AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]

Applying the coordinates 2,6 and -2 to the basis E we obtain

2+6t-2t^{2}

That was the original result of T[e1(t)]

8 0
3 years ago
I need help with this question
kifflom [539]

Answer:

(-1,9)

(0,3)

(1,-3)

(5,-27)

Step-by-step explanation:

Plug x values into the equation.

6 0
2 years ago
X = 18 is a solution to this equation X + 79 = 98 Question 4 options: True False
deff fn [24]
X+79=98 (subtract 79 to isolate x)
X=19 so it is false
7 0
3 years ago
Read 2 more answers
Which of the following square root of -80
Marta_Voda [28]

You can factor -80 as

-80 = (-1)\cdot 16 \cdot 5

So, we have

\sqrt{-80} = \sqrt{(-1)\cdot 16 \cdot 5}

The square root of a product is the product of the square roots:

\sqrt{-80} = \sqrt{(-1)}\sqrt{16}\sqrt{5}

Since i^2=-1 and 4^2=16, we have

\sqrt{-80} = 4i\sqrt{5}

8 0
3 years ago
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