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3 years ago

How many solutions does the system of linear equations 6a + 3b = 12 and 6a - 3b = 12 have?

Mathematics

1 answer:

Olin [163]3 years ago

3 0

Answer:

One solution

Step-by-step explanation:

Lets solve it for b

3b= -6a+12

3b= 6a-12

Simplifying

b= -2a+4

b= 2a+4

Both have same slope and intercept hence both lines are going to intersect at one point. So the equations are having one solution.

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AB=4x+3 and CD=24+5x find bc

madam [21]

Answer:

Step-by-step explanation:

1. use distribution factor

(4x+3)(24+5x)

4x times 24 = 96

4x times 5x = 20x

3 times 24 = 72

3 times 5x = 15x

2. combine like terms

96 + 72 = 168

15x + 20x = 35x

35x + 168 or 168 + 35x

I may be wrong please check my answer. I may have a lapse in my algebra, or I may have misunderstood the problem. But this is what I can give you to the best of my knowledge.

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3 years ago

3^2 + (6 • 4^2)10 - 22

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9+(<span>6•16)10-22
9+(96)10-22
9+960-22
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3 years ago

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3 years ago

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Which expression has a quotient of about 7

Naily [24]

Answer:

H (36 / 5)

Divide each of the numbers and see which one is closest to seven

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2 years ago

Solve the equation on the interval [0,2π]

WARRIOR [948]

$\bf 16sin^5(x)+2sin(x)=12sin^3(x)
\\\\\\
16sin^5(x)+2sin(x)-12sin^3(x)=0
\\\\\\
\stackrel{common~factor}{2sin(x)}[8sin^4(x)+1-6sin^2(x)]=0\\\\
-------------------------------\\\\
2sin(x)=0\implies sin(x)=0\implies \measuredangle x=sin^{-1}(0)\implies \measuredangle x=
\begin{cases}
0\\
\pi \\
2\pi 
\end{cases}\\\\
-------------------------------$

$\bf 8sin^4(x)+1-6sin^2(x)=0\implies 8sin^4(x)-6sin^2(x)+1=0$

now, this is a quadratic equation, but the roots do not come out as integers, however it does have them, the discriminant, b² - 4ac, is positive, so it has 2 roots, so we'll plug it in the quadratic formula,

$\bf 8sin^4(x)-6sin^2(x)+1=0\implies 8[~[sin(x)]^2~]^2-6[sin(x)]^2+1=0
\\\\\\
~~~~~~~~~~~~\textit{quadratic formula}
\\\\
\begin{array}{lcccl}
& 8 sin^4& -6 sin^2(x)& +1\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array} 
\qquad \qquad 
sin(x)= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}
\\\\\\
sin(x)=\cfrac{-(-6)\pm\sqrt{(-6)^2-4(8)(1)}}{2(8)}\implies sin(x)=\cfrac{6\pm\sqrt{4}}{16}
\\\\\\
sin(x)=\cfrac{6\pm 2}{16}\implies sin(x)=
\begin{cases}
\frac{1}{2}\\\\
\frac{1}{4}
\end{cases}$

$\bf \measuredangle x=
\begin{cases}
sin^{-1}\left( \frac{1}{2} \right)
sin^{-1}\left( \frac{1}{4} \right)
\end{cases}\implies \measuredangle x=
\begin{cases}
\frac{\pi }{6}~,~\frac{5\pi }{6}\\
----------\\
\approx~0.252680~radians\\
\qquad or\\
\approx~14.47751~de grees\\
----------\\
\pi -0.252680\\
\approx 2.88891~radians\\
\qquad or\\
180-14.47751\\
\approx 165.52249~de grees
\end{cases}$

3 0

3 years ago