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3 years ago

Which real value of m makes the expression undefined M^8+1/4m^2

Mathematics

1 answer:

Soloha48 [4]3 years ago

5 0

Answer:

m = 0

Step-by-step explanation:

The denominator of the rational expression cannot be zero as this would make the rational expression undefined. Equating the denominator to zero and solving gives the value that m cannot be.

4m² = 0 ⇒ m² = 0 ⇒ m = 0 ← excluded value

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Which of the following are solutions to the equation below?

sladkih [1.3K]

Answer:

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

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Step-by-step explanation:

considering the equation

$2x^2\:-\:4x\:-\:3\:=\:x$

solving

$2x^2\:-\:4x\:-\:3\:=\:x$

$2x^2-4x-3-x=x-x$

$2x^2-5x-3=0$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=-5,\:c=-3:\quad x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

solving

$x=\frac{-\left(-5\right)+\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$x=\frac{5+\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$x=\frac{5+\sqrt{49}}{2\cdot \:2}$

$x=\frac{5+7}{4}$

$x=3$

also solving

$x=\frac{-\left(-5\right)-\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$x=\frac{5-\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$x=\frac{5-\sqrt{49}}{4}$

$x=-\frac{2}{4}$

$x=-\frac{1}{2}$

Therefore,

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=3,\:x=-\frac{1}{2}$

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3 years ago

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Given:

The width at the base of parabolic tunnel is 7 m.

The ceiling 3 m from each end of the base there are light fixtures.

The height to light fixtures is 4 m.

To find:

Whether it is possible a trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel.

Solution:

The width at the base of tunnel is 7 m.

Let the graph of the parabola intersect the x-axis at x=0 and x=7. It means x and (x-7) are the factors of the height function.

The function of height is:

$h(x)=ax(x-7)$ ...(i)

Where, a is a constant.

The ceiling 3 m from each end of the base there are light fixtures and the height to light fixtures is 4 m. It means the graph of height function passes through the point (3,4).

Putting x=3 and h(x)=4 in (i), we get

$4=a(3)((3)-7)$

$4=a(3)(-4)$

$\dfrac{4}{(3)(-4)}=a$

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Putting $a=-\dfrac{1}{3}$ , we get

$h(x)=-\dfrac{1}{3}x(x-7)$ ...(ii)

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The width of the truck is 4 m. If is passes through the center then the truck must m 2 m on the left side of the center and 2 m on the right side of the center.

2 m on the left side of the center is x=1.5.

A trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel is possible if h(1.5) is greater than 2.8.

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$h(1.5)=-\dfrac{1}{3}(1.5)(1.5-7)$

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Since h(1.5)<2.8, therefore the trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel is not possible.

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