I need answer pleaseee

Fatima writes down three square numbers.

nekit [7.7K]

Answer:

10

Step-by-step explanation:

Fatima wrote down 3 square numbers

since we don't know the three square we will represent it by any variable (X)

3x=30

so you divide both sides by 3

so we have 3x/3=30/3=10

therefore the answer is 10

8 0

3 years ago

Please answer this I don't understand at all

katrin [286]

This is your answer, i hope this help if you have any questions let me know

8 0

3 years ago

Helpppppppppp me plzzzzzzzzzz WILL MARK BRAINLIST if right

Dvinal [7]

Step-by-step explanation:

j=-m+g that's it I just get it

6 0

3 years ago

Find a decomposition of a=⟨−5,−1,1⟩ into a vector c parallel to b=⟨−6,0,6⟩ and a vector d perpendicular to b such that c+d=a.

dezoksy [38]

The projection of vector A parallel to vector B is $\langle -3, 0, 3\rangle$ and the projection of vector A perpendicular to vector B is $\langle -2, -1, -2\rangle$ .

In this question, we need to determine all projections of a vector with respect to another vector. In this case, the projection of vector A parallel to vector B is defined by this formula:

$\vec a_{\parallel , \vec b} = \frac{\vec a \,\bullet\,\vec b}{\|\vec b\|^{2}}\cdot \vec b$ (1)

Where $\|\vec b\|$ is the norm of vector B.

And the projection of vector A perpendicular to vector B is:

$\vec a_{\perp, \vec b} = \vec a - \vec a_{\parallel, \vec b}$ (2)

If we know that $a = \langle -5, -1, 1 \rangle$ and $\vec b = \langle -6, 0, 6 \rangle$ , then the projections are now calculated:

$\vec a_{\parallel, \vec b} = \frac{(-5)\cdot (-6)+(-1)\cdot (0)+(1)\cdot (6)}{(-6)^{2}+0^{2}+6^{2}} \cdot \langle -6, 0, 6 \rangle$

$\vec a_{\parallel, \vec b} = \frac{1}{2}\cdot \langle -6, 0, 6 \rangle$

$\vec a_{\parallel, \vec b} = \langle -3, 0, 3\rangle$

$\vec a_{\perp, \vec b} = \langle -5, -1, 1 \rangle - \langle -3, 0, 3 \rangle$

$\vec a_{\perp, \vec b} = \langle -2, -1, -2\rangle$

The projection of vector A parallel to vector B is $\langle -3, 0, 3\rangle$ and the projection of vector A perpendicular to vector B is $\langle -2, -1, -2\rangle$ .

We kindly invite to check this question on projection of vectors: brainly.com/question/24160729

7 0

3 years ago

If the 13th and 38th terms of an arithmetic sequence are -53 and -128,

love history [14]

9514 1404 393

Answer:

-2018

Step-by-step explanation:

The n-th term is ...

an = a1 +d(n -1)

So, the given terms are ...

-53 = a1 +12d

-128 = a1 +37d

Subtracting the first from the second gives ...

(a1 +37d) -(a1 +12d) = (-128) -(-53)

25d = -75

d = -3

The 668th term will be ...

a668 = a1 +d(668 -1) = a1 +667d = (a1 +37d) +630d

a668 = -128 +630(-3) = -128 -1890 . . . . substitute for a38

a668 = -2018

7 0

3 years ago