What is the equation of a line that is parallel to the line y = 2x+1 and passes

18/6 equivalent ratio

Anon25 [30]

Answer:

3/1 ; 9/ 3 ; 36/ 18

Step-by-step explanation:

Write the ratios in fractional form

Initial condition:

18

6

1

st Equivalent

→

18

÷

6

÷

6

=

3

1

2

nd

Equivalent

→

18

÷

2

6

÷

2

=

9

3

rd

Equivalent

→

18

×

2

6

×

2

=

36

12

Write the ratios in fractional form

Initial condition:

18

6

1

st

Equivalent

→

18

÷

6

÷

6

=

3

1

2

nd

Equivalent

→

18

÷

2

6

÷

2

=

9

3

rd

Equivalent

→

18

×

2

6

×

2

=

36

12

4 0

3 years ago

Read 2 more answers

Otis had a salad for $4.50, a sandwich for $6.25, and a drink at his favorite restaurant. The tax for the entire meal was $0.54.

UkoKoshka [18]

Assuming 4% tax.

(4.50+6.25+d)0.04= 0.54

(10.75+d)=13.5

d= $2.75

3 0

3 years ago

I NEED HELP ASAP PLEASEEE!!!! WILL GIVE BRAINLIEST !!

scoray [572]

Answer:

it's d credit to the people who said it

6 0

3 years ago

Consider f(x) = 1.8x – 10 and g(x) = −4. Select the equation that can be used to find the input value at which f (x ) = g (x ),

serg [7]

Answer:

Option A- 1.8x – 10 = –4; x = 1.8 x minus 10 equals negative 4; x equals StartFraction 10 Over 2 EndFraction.

3 0

4 years ago

Read 2 more answers

Find a formula for the least squares solution of ax=b when the columns of a are orthonormal1 .

neonofarm [45]

Let $\mathbf A$ be a rectangular $m\times n$ matrix with column vectors $\mathbf a_1,\ldots,\mathbf a_n$ , i.e.

$\mathbf A=\begin{bmatrix}\mathbf a_1&\cdots&\mathbf a_n\end{bmatrix}$

Then we have

$\mathbf A^\top=\begin{bmatrix}\mathbf a_1&\cdots&\mathbf a_n\end{bmatrix}^\top$

and the product of the two is

$\mathbf A^\top\mathbf A=\begin{bmatrix}\mathbf a_1\cdot\mathbf a_1&\mathbf a_1\cdot\mathbf a_2&\cdots&\mathbf a_1\cdot\mathbf a_n\\\mathbf a_2\cdot\mathbf a_1&\mathbf a_2\cdot\mathbf a_2&\cdots&\mathbf a_2\cdot\mathbf a_n\\\vdots&\vdots&\ddots&\vdots\\\mathbf a_n\cdot\mathbf a_1&\mathbf a_n\cdot\mathbf a_2&\cdots&\mathbf a_n\cdot\mathbf a_n\end{bmatrix}$

Because the columns of $\mathbf A$ are orthonormal, we have

$\mathbf a_i\cdot\mathbf a_j=\begin{cases}1&\text{for }i=j\\0&\text{for }i\neq j\end{cases}$

which means $\mathbf A^\top\mathbf A$ reduces to an $n\times n$ matrix with ones along the diagonal and zero everywhere else, i.e.

$\mathbf A^\top\mathbf A=\begin{bmatrix}1&0&\cdots&0\\0&1&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&1\end{bmatrix}=\mathbf I_n$

where $\mathbf I$ denotes the identity matrix. This means the solution to $\mathbf{Ax}=\mathbf b$ is given by

$\mathbf A^\top(\mathbf{Ax})=\mathbf A^\top\mathbf b\implies(\underbrace{\mathbf A^\top\mathbf A}_{\mathbf I})\mathbf x=\mathbf A^\top\mathbf b\implies\mathbf x=\mathbf A^\top\mathbf b$

6 0

3 years ago