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saveliy_v [14]
3 years ago
11

I need help what’s the answer

Mathematics
1 answer:
alexira [117]3 years ago
7 0

Answer:

The square of a binomial pattern

Step-by-step explanation:

Given

9x² - 30x + 25

Consider the factors of the product of the coefficient of the x² term and the constant term which sun to give the coefficient of the x- term.

product = 9 × 25 = 225 and sum = - 30

The factors are - 15 and - 15

Use these factors to split the x- term

9x² - 15x - 15x + 25 ( factor the first/second and third/fourth terms )

= 3x(3x - 5) - 5(3x - 5) ← factor out (3x - 5) from each term

= (3x - 5)(3x - 5)

= (3x - 5)² ← the square of a binomial pattern

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Jamie went on a vacation with his family. He spent 32.25% of the total cost on the airline tickets, of the total cost on hotels,
aniked [119]

Answer:

In my opinion he spent 55%

7 0
3 years ago
30 Points if you can answer my question
Leviafan [203]

Answer:


Step-by-step explanation:

6x=3(x+4) distribute the 3 to the x and the 4 in the parenthesis to get 6x=3x+12 then subtract 3x on both sides to get 3x=12 last step is to divide by 3 on both sides so x is by itself and all you have left is 12/3 which would leave you with x=4



5 0
3 years ago
A contractor is building a set of stairs out of concrete. Each stair is exactly the same length, width, and height.
Tpy6a [65]

Answer:

Dimensions:

H = 0.9 m

W = 0.4 M

L = 2.4 m

Step-by-step explanation:

As they are all the same dimensions and there are only 3 stairs, divide all the figures (except the length as it will stay the same) by 3 to get your figures.

3 0
3 years ago
Read 2 more answers
If A = 1 2 1 1 and B= 0 -1 1 2 then show that (AB)^-1 = B^-1 A^-1<br><br><br> help meeeee plessss ​
Trava [24]

A = \begin{bmatrix}1&2\\1&1\end{bmatrix} \implies A^{-1} = \dfrac1{\det(A)}\begin{bmatrix}1&-1\\-2&1\end{bmatrix} = \begin{bmatrix}-1&1\\2&-1\end{bmatrix}

where det(<em>A</em>) = 1×1 - 2×1 = -1.

B = \begin{bmatrix}0&-1\\1&2\end{bmatrix} \implies B^{-1} = \dfrac1{\det(B)}\begin{bmatrix}2&1\\-1&0\end{bmatrix} = \begin{bmatrix}2&1\\-1&0\end{bmatrix}

where det(<em>B</em>) = 0×2 - (-1)×1 = 1. Then

B^{-1}A^{-1} = \begin{bmatrix}2&1\\-1&0\end{bmatrix} \begin{bmatrix}-1&1\\2&-1\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}

On the other side, we have

AB = \begin{bmatrix}1&2\\1&1\end{bmatrix} \begin{bmatrix}0&-1\\1&2\end{bmatrix} = \begin{bmatrix}2&3\\1&1\end{bmatrix}

and det(<em>AB</em>) = det(<em>A</em>) det(<em>B</em>) = (-1)×1 = -1. So

(AB)^{-1} = \dfrac1{\det(AB)}\begin{bmatrix}1&-3\\-1&2\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}

and both matrices are clearly the same.

More generally, we have by definition of inverse,

(AB)(AB)^{-1} = I

where I is the identity matrix. Multiply on the left by <em>A </em>⁻¹ to get

A^{-1}(AB)(AB)^{-1} = A^{-1}I = A^{-1}

Multiplication of matrices is associative, so we can regroup terms as

(A^{-1}A)B(AB)^{-1} = A^{-1} \\\\ B(AB)^{-1} = A^{-1}

Now multiply again on the left by <em>B</em> ⁻¹ and do the same thing:

B^{-1}\left(B(AB)^{-1}\right) = (B^{-1}B)(AB)^{-1} = B^{-1}A^{-1} \\\\ (AB)^{-1} = B^{-1}A^{-1}

7 0
3 years ago
Can u help??????????????????????????????????????//
maw [93]

Answer:

Okay!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

The answer is 14

Step-by-step explanation:

Folow order of operations and when doing multipactiona or subtraction or division and addition, make sure to go from left to right

5 0
3 years ago
Read 2 more answers
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