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3 years ago

Find quotient using long division

Mathematics

2 answers:

Andrei [34K]3 years ago

7 0

Answer: 3x+5+(21)/x-4

Step-by-step explanation:

Nezavi [6.7K]3 years ago

4 0

Answer:3x+5+(21)/x-4

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Solve for x. (1 point) −3x + 2b > 8 x < the quantity negative 2 times b plus 8 all over negative 3 x > the quantity neg

masha68 [24]

Given:

$-3x+2b>8$

To find:

The value of x.

Solution:

We have,

$-3x+2b>8$

To find the value of x, we need to isolate x on one side.

Subtract 2b from both sides.

$-3x>8-2b$

Divide both sides by -3. On multiplying or dividing an inequality by a negative number, we need to change the sign of inequality.

$\dfrac{-3x}{-3}$

$x$

The required inequality for x is $x$ .

Therefore, the correct option is A.

8 0

4 years ago

What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$