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olga2289 [7]
3 years ago
6

What is -4d 3 + 28d 2 - 4d

Mathematics
1 answer:
Svetlanka [38]3 years ago
7 0

Answer: 20d + 5

Step-by-step explanation:

First we need to combine like terms.

In this equation we can see that they're are 3 numbers containing the letter D we have -4, -4, 28.

We're going to combine -4 and -4 to get -8 and then combine -8 to 28 getting 20d

Afterwards we're going to combine 3, 2 equaling 5.

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dexar [7]

Answer:

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Step-by-step explanation:

3 0
3 years ago
What is the cube root of 27a^12
fiasKO [112]
Cube root of 27 a^12 will be:  3 a^4.
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3 years ago
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Please answer this question :))<br><br> 2mn^2 (3m^2 + 5n)
Andreyy89

Answer:

23

Step-by-step explanation:

3 0
3 years ago
Jessica and Josh are selling Entertainment Books to raise money for the art room at their school. Jessica received the prize for
Over [174]

Answer:

Josh sold 16 books and Jessica sold 240 books

Step-by-step explanation:

Given: Jessica sold 15 times more books than Josh.

           Together they sold 256 books

To Find : How many did each one of them sell?

Solution:

Let Jessica sold x books

Let Josh sold y books.

Since we are given that Jessica sold 15 times more books than Josh.

⇒x=15y --a

We are also given that  Together they sold 256 books.

⇒x+y=256 --b

Using substitution method

Put value of x from equation a in equation b

⇒15y+y=256

⇒16y=256

⇒y=\frac{256}{16}

⇒y=16

Thus Josh sold 16 books

Jessica sold books = 15y = 15*16=240

Hence  Josh sold 16 books and Jessica sold 240 books

8 0
3 years ago
Read 2 more answers
13. Determine whether B = {(-1, 1,-1), (1, 0, 2), (1, 1, 0)} is a basis of R3.
Sholpan [36]

Answer:  Yes, the given set of vectors is a basis of R³.

Step-by-step explanation:  We are given to determine whether the following set of three vectors in R³ is a basis of R³ or not :

B = {(-1, 1,-1), (1, 0, 2), (1, 1, 0)} .

For a set to be a basis of R³, the following two conditions must be fulfilled :

(i) The set should contain three vectors, equal to the dimension of R³

and

(ii) the three vectors must be linearly independent.

The first condition is already fulfilled since we have three vectors in set B.

Now, to check the independence, we will find the determinant formed by theses three vectors as rows.

If the value of the determinant is non zero, then the vectors are linearly independent.

The value of the determinant can be found as follows :

D\\\\\\=\begin{vmatrix} -1& 1 & -1\\ 1 & 0 & 2\\ 1 & 1 & 0\end{vmatrix}\\\\\\=-1(0\times0-2\times1)+1(2\times1-1\times0)-1(1\times1-0\times1)\\\\=(-1)\times(-2)+1\times2-1\times1\\\\=2+2-1\\\\=3\neq 0.

Therefore, the determinant is not equal to 0 and so the given set of vectors is linearly independent.

Thus, the given set is a basis of R³.

5 0
3 years ago
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