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ohaa [14]
3 years ago
7

The polynomial x3 + 64 is an example of a

Mathematics
1 answer:
liubo4ka [24]3 years ago
6 0

Because the value of x is raised to the power of 3, x^3 is a cubed number.

and 64 is 4^3

This makes the polynomial a sum of cubes.

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Which figure represents the image of parallelogram LMNP after a reflection across the line y = x?
Taya2010 [7]
When you reflect a point across the line y = x, the x-coordinate and y-coordinate change places. This gives you such reflection rule:

From the diagram:
L(3,1), M(4,3), N(5,3) and P(4,1).
Using the reflection rule, you can find coordinates of image points:
L'(1,3), M'(3,4), N'(3,5) and P'(1,4).
As you can see, these are coordinates of vertices of the figure C.
Answer: correct choice is option 3 - figure C.
7 0
4 years ago
A student spends 12% of his or her money on a book. After buying the book, the student has $120.12 left.
dezoksy [38]

Answer:

$136.50

Step-by-step explanation:

100 - 12 = 88% (How much percentage of money the student left)

88% = 120.12

1% = 120.12 ÷ 88 = 1.365

Before buying the book, the student had 100%

100% = 1.365 x 100 = 136.50

5 0
3 years ago
Can someone plz help me with this one problem plz!!!
Xelga [282]
The area is 18 square miles
7 0
3 years ago
Which of the following values is a solution to the inequality? 12 – 5x > -8
olya-2409 [2.1K]

Answer:

x < 4

Step-by-step explanation:

12 – 5x > -8

Subtract 12 from each side

12-12 – 5x > -8-12

-5x> -20

Divide each side by -5, remembering to flip the inequality

-5x/-5 < -20/-5

x < 4

5 0
3 years ago
What is the 4th term of the expanded binomial (2x – 3y)^6
san4es73 [151]

Answer:

The 4th term of the expanded binomial is -4320x^3y^3

Step-by-step explanation:

Considering:

$ (x+y)^n = \sum_{k=0}^{n} \binom{n}{k}  x^{n-k}y^k$

$ (2x-3y)^6 = \sum_{k=0}^{6} \binom{6}{k}  (2x)^{6-k}(-3y)^k$

Now, you gotta calculate for every value of k

$ (2x-3y)^6 = \binom{6}{0}  (2x)^{6-0}(-3y)^0     +       \binom{6}{1}  (2x)^{6-1}(-3y)^1     +      \binom{6}{2}  (2x)^{6-2}(-3y)^2   +   \\ $

$\binom{6}{3}  (2x)^{6-3}(-3y)^3    +    \binom{6}{4}  (2x)^{6-4}(-3y)^4    +  \binom{6}{5}  (2x)^{6-5}(-3y)^5    +    \binom{6}{6}  (2x)^{6-6}(-3y)^6            $

I will not write every product, but just solve following the steps:

For k=0

$\binom{6}{0}  (2x)^{6-0}(-3y)^0$

$\frac{6!}{(6-0)!(0!)}   (2x)^{6-0}(-3y)^0$

$ \frac{6!}{6!} \left(2x\right)^{6-0}\cdot 1$

$1\cdot \:1\cdot \left(2x\right)^{6-0}$

$2^6x^6$

64x^6

(2x-3y)^6=64x^6-576x^5y+2160x^4y^2-4320x^3y^3+4860x^2y^4-2916xy^5+729y^6

8 0
3 years ago
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