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3 years ago

PLEASE HELP!!! Describe the graph for: y< 4x + 3

Mathematics

2 answers:

ELEN [110]3 years ago

7 0

The equation show that it is a linear graph this means the graph is a straight line

den301095 [7]3 years ago

7 0

The answer is -1x & 3y

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<img src="https://tex.z-dn.net/?f=3x%20-%20%285x%20-%209%29%20%3D%207%20-%204%28x%20%2B%201%29" id="TexFormula1" title="3x - (5x

Montano1993 [528]

$3x-(5x-9)=7-4(x+1)\\\\3x-5x+9=7-4x-4\\\\-2x+9=-4x+3\ \ \ \ |-9\\\\-2x=-4x-6\ \ \ \ |+4x\\\\2x=-6\ \ \ \ |:2\\\\x=-3$

8 0

3 years ago

70 POINTS QUESTION ATTACHED (actually attached this time lol)

omeli [17]

Answer:

(1/5 h)^2

Step-by-step explanation:

1/25 h^2

To make this a perfect square we need to take the square root and then square it

(sqrt(1/25 h^2))^2

we know that sqrt(ab) = sqrt(a) sqrt(b)

(sqrt(1/25) sqrt(h^2))^2

(1/5 h)^2

6 0

3 years ago

Read 2 more answers

The double box plot shows the test scores for Miss Robinson's second and fourth period

Afina-wow [57]

Answer: The second period class has the greater first quartile.

4 0

4 years ago

mr. Chaplin measure distance rectangular bedroom for new carpeting who's bedrooms was the greatest area of Georgia's better and

Alina [70]

4.4 is the largest so Georgia's is better

6 0

3 years ago

For the following linear system, put the augmented coefficient matrix into reduced row-echelon form.

Anni [7]

Answer:

The reduced row-echelon form of the linear system is $\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]$

Step-by-step explanation:

We will solve the original system of linear equations by performing a sequence of the following elementary row operations on the augmented matrix:

Interchange two rows
Multiply one row by a nonzero number
Add a multiple of one row to a different row

To find the reduced row-echelon form of this augmented matrix

$\left[\begin{array}{cccc}2&3&-1&14\\1&2&1&4\\5&9&2&7\end{array}\right]$

You need to follow these steps:

Divide row 1 by 2 $\left(R_1=\frac{R_1}{2}\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\1&2&1&4\\5&9&2&7\end{array}\right]$

Subtract row 1 from row 2 $\left(R_2=R_2-R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\5&9&2&7\end{array}\right]$

Subtract row 1 multiplied by 5 from row 3 $\left(R_3=R_3-\left(5\right)R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 1 $\left(R_1=R_1-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 3 $\left(R_3=R_3-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&0&0&-19\end{array}\right]$

Multiply row 2 by 2 $\left(R_2=\left(2\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&-19\end{array}\right]$

Divide row 3 by −19 $\left(R_3=\frac{R_3}{-19}\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&1\end{array}\right]$

Subtract row 3 multiplied by 16 from row 1 $\left(R_1=R_1-\left(16\right)R_3\right)$

$\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&-6\\0&0&0&1\end{array}\right]$

Add row 3 multiplied by 6 to row 2 $\left(R_2=R_2+\left(6\right)R_3\right)$

$\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]$

8 0

3 years ago