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2 years ago

16 divided by 1.296 Can you take me through step-by-step on this plz?

Mathematics

2 answers:

eduard2 years ago

3 0

Answer:

12.3456790123

Step-by-step explanation:

its hard to explain but you could always use a clculator and no I did not use one. But these are hard I'm a math major

Alexxandr [17]2 years ago

3 0

1 2 . 3 4 5 6 7 9

1 2 9 6 | 1 6 0 0 0 .

1 2 9 6

3 0 4 0 .

2 5 9 2 .

4 4 8 . 0

3 8 8 . 8

5 9 . 2 0

5 1 . 8 4

7 . 3 6 0

6 . 4 8 0

8 8 0 0

7 7 7 6

1 0 2 4 0

9 0 7 2

1 1 6 8 0

1 1 6 6 4

1 6

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dybincka [34]

16 ...........jsjshshsh

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3 years ago

How is the graph of y=9(3)^x-2 -6 translated from the graph of y=9(3)^x ?

USPshnik [31]

For this case we have the following function:
y = 9 (3) ^ x
Applying the following transformations we have:
Horizontal translations
Suppose that h> 0
To graph y = f (x-h), move the graph of h units to the right.
y = 9 (3) ^ (x-2)
Vertical translations
Suppose that k> 0
To graph y = f (x) -k, move the graph of k units down.
y = 9 (3) ^ (x-2) - 6
Answer:
2 units to the right
6 units down

5 0

3 years ago

Solve the inequality 4t^2 ≤ 9t-2 please show steps and interval notation. thank you!

FromTheMoon [43]

Answer: $t\in [\dfrac{1}{4},2]$

Step-by-step explanation:

Given

Inequality is $4t^2\leq9t-2$

Taking variables one side

$\Rightarrow 4t^2-9t+2\leq0\\\Rightarrow 4t^2-8t-t+2\leq0\\\Rightarrow 4t(t-2)-1(t-2)\leq0\\\Rightarrow (4t-1)(t-2)\leq0$

Using wavy curve method

$t\in [\dfrac{1}{4},2]$

3 0

2 years ago

Question number 16 pls answer

lozanna [386]

3x - 6 = 4x + 1
3x - 6 + 6 = 4x + 1 + 6
3x = 4x + 7
3x - 4x = 4x + 7 - 4x

X = -7
:)

5 0

2 years ago

Read 2 more answers

Find the inverse of ????−7 −2

Masja [62]

Answer:

$A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

Step-by-step explanation:

$A=\begin{bmatrix}-7 & -2\\ 4 & 1\end{bmatrix}$

$det\ A=-7\times 1-(-2)\times 4\\\Rightarrow det\ A=1$

$A^{-1}=\frac{1}{det\ A}\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A^{-1}=\frac{1}{1}\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

$\therefore A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

$A.A^{-1}=\begin{bmatrix}-7 & -2\\ 4 & 1\end{bmatrix}\times \begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}\left(-7\right)\cdot \:1+\left(-2\right)\left(-4\right)&\left(-7\right)\cdot \:2+\left(-2\right)\left(-7\right)\\ 4\cdot \:1+1\cdot \left(-4\right)&4\cdot \:2+1\cdot \left(-7\right)\end{pmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$

Hence, confirmed

6 0

3 years ago