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2 years ago

Mszbnxcvlhjuascv,hjaqsc,hjkas

Mathematics

2 answers:

Wittaler [7]2 years ago

6 0

Answer:

Step-by-step explanation:

A is a parabola

ELEN [110]2 years ago

5 0

A
Not sure thoooooooooo

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4 0

3 years ago

FAST

sertanlavr [38]

Answer: The system consists of parallel lines

Step-by-step explanation:

Given system of lines :

$y=\dfrac13x-4$

$3y-x=-7$

Substitute $y=\dfrac13x-4$ in $3y-x=-7$ , we get

$3(\dfrac13x-4)-x=-7\\\\\Rightarrow\ x-12-x=-7\\\\\Rightarrow\ -12=-7$ , which is not true.

That means , system has no solution.

i.e. they are representing parallel lines. [Parallel lines do not intersect and hence they do not have solution.]

6 0

2 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,

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