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3 years ago

15

Find the indicated products. (3x-5)(3x+5)

2 answers:

frutty [35]3 years ago

8 0

9x²- 25

3x(3x) = 9x²
3x(5) = 15x
-5(3x) = -15x
-5(5) = -25
9x² + 15x - 15x - 25
9x² - 25 is your answer

hope this helps

11111nata11111 [884]3 years ago

4 0

The answer would be 9x to the power of 2 - 25 because u multiply 3 and 3 and 5 and 5

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Help please!!!!!!!!!

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ANSWER

24

EXPLANATION

For a matrix A of order n×n, the cofactor $C_{ij}$ of element $a_{ij}$ is defined to be

$C_{ij} = (-1)^{i+j} M_{ij}$

$M_{ij}$ is the minor of element $a_{ij}$ equal to the determinant of the matrix we get by taking matrix A and deleting row i and column j.

Here, we have

$C_{11} = (-1)^{1+1} M_{11} = M_{11}$

M₁₁ is the determinant of the matrix that is matrix A with row 1 and column 1 removed. The bold entries are the row and the column we delete.

$\begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \end{aligned}$

Since the determinant of a 2×2 matrix is

$\det\left( \begin{bmatrix} a & b \\ c& d \end{bmatrix} \right) = ad-bc$

it follows that

$\begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \\ &= (0)(-8) - (-3)(8) \\ &= -(-24) \\ &= 24 \end{aligned}$

so $C_{11} = M_{11} = 24$

4 0

3 years ago

Round 479.8983 to the nearest whole number:

Slav-nsk [51]

The answer to your question is 480

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3 years ago

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Determine the temperature of 2.6 moles of gas contained in a 5.00-L vessel at a pressure of 1.2atm.

strojnjashka [21]

Answer:

28.108 K.

Step-by-step explanation:

Given: Pressure (P)= 1.2atm

Number of moles (n)= 2.6 moles

Volume (V)= 5.00-L

Now finding the temperature (T).

Formula; T= $\frac{P\times V}{n\times R}$

R is a constant factor which makes other factors work together.

There is a numerical value for R which we use is $0.0821 \times \frac{L.atm}{mole.K}$

∴ Temperature (T)= $\frac{1.2\times 5}{2.6\times 0.0821 \frac{L.atm}{mol.K} }$

⇒ Temperature (T)= $\frac{6}{0.21346} = 28.1083\ K$

∴ Temperature is 28.108 K

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shepuryov [24]

Answer:

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Step-by-step explanation:

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