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sveticcg [70]
3 years ago
15

Which polynomial represents the algebra tile configuration shown? x2 3x 4 x2 3x â’ 4x â’x2 â’ 7 â’x2 3x â’ 4.

Mathematics
1 answer:
Ilya [14]3 years ago
8 0

The polynomial which shows algebra tile configuration is \rm -x^2 + 3x - 4.

We have to determine

Which polynomial represents the algebra tile configuration shown?

According to the question

Polynomials are defined as the sums of terms that are represented in the form of k-xⁿ, where k is any number and n is a positive integer.

The polynomial represented by algebra tiles is \rm -x^2 + 3x - 4

In the given tiles,

2 tiles for = \rm -x^2

3 tiles for = x

4 tiles for = -4

Now, putting the like terms together, the equation becomes:

\rm (x^2+-2x^2) + 3(x+  x+ x ) -(1+1+1+ 1)

The like terms can be added such that:

\rm -x^2 + 3x - 4

Hence, the polynomial which shows algebra tile configuration is \rm -x^2 + 3x - 4.

To know more about Polynomial click the link given below.

brainly.com/question/26080405

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Write the equation of a circle with a center at (-2, 3) and a radius of 4.
Ivan

Answer:

The equation of a circle:

(x+2)^2+(y-3)^2 = 16

Step-by-step explanation:

The equation of a circle:

(x-h)^2+(y-k)^2 = r^2

(where the center is known as (h,k) and the radius, known as r).

-Since you already have the center and the radius, use them for the equation of a circle:

Center: (-2,3)

Radius: 4

(x+2)^2+(y-3)^2 = 4^2

After you have the equation of a circle, simplify the radius by the exponent 2 :

(x+2)^2+(y-3)^2 = 16

So, you have found the equation of a circle.

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8 0
3 years ago
Find a cubic function f(x) = ax3 + bx2 + cx + d that has a local maximum value of 3 at x = −3 and a local minimum value of 0 at
Amiraneli [1.4K]

Answer:

The cubic function is f(x) = (27/32)·x  - 3/32·x³ - -9/32·x² - 9/32

Step-by-step explanation:

The given function is f(x) = a·x³ + b·x² + c·x + d

By differentiation, we have;

3·a·x² + 2·b·x + c = 0

3·a·(-3)² + 2·b·(-3) + c = 0

3·a·9 - 6·b + c = 0

27·a - 6·b + c = 0

3·a·(1)² + 2·b·(1) + c = 0

3·a + 2·b + c = 0

a·(-3)³ + b·(-3)² + c·(-3) + d = -3

-27·a + 9·b - 3·c + d = -3...(1)

a + b + c + d = 0...(2)

Subtracting equation (1) from equation (2) gives;

28·a - 8·b + 4·c = 3

Therefore, we have;

27·a - 6·b + c = 0

3·a + 2·b + c = 0

28·a - 8·b + 4·c = 0

Solving the system of equations using an Wolfram Alpha gives;

a = -3/32, b = -9/32, c = 27/32 from which we have;

a + b + c + d = 0 3 × (-3/32) + 2 × (-9/32) + (27/32) + d = 0

d = 0 - (0 3 × (-3/32) + 2 × (-9/32) + (27/32)) = -9/32

The cubic function is therefore f(x) = (-3/32)·x³ + (-9/32)·x² + (27/32)·x + (-9/32).

4 0
3 years ago
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