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2 years ago

Which values of c will cause the quadratic equation –x2 + 3x + c = 0 to have no real number solutions? Check all that apply.

2 answers:

7 0

Quadratic equation which has no solution gives zero for the value of d.
d = b² - 4ac
0 = 3² +- 4(-1)c
-9 = +_ 4c
c = -9/-4 pr 9/-4
c = 9/4 or -9/4

In short, Your Answers would be Option C & E

Hope this helps!

Lerok [7]2 years ago

7 0

Answer:

the correct answer is A and B

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a pyramid whose altitude is 5ft weighs 800lbs. at what distance from its vertex must it be cut by a plane parallel to its base s

slavikrds [6]

A structure which has a square base and four triangular sides meeting at a point is called pyramid.

At distance of 3.97 feet from its vertex , pyramid is cut by plane so that the two solids of equal weight will be formed.

<u>It is assumed that weight of pyramid is proportional to its volume.</u>

So, $w = k V$ , where V is volume of original pyramid and v is volume of small pyramid and k is constant.

Let us consider that at h distance, pyramid is cut from its vertex. So a small pyramid is also formed.

Assume that base area of original pyramid is A and base of small pyramid is a.

Volume of original pyramid is, $V= \frac{1}{3} *A* 5$

So, weight of original pyramid, $W = k *(\frac{1}{3} *A* 5) = 800$

Volume of small pyramid is, $v = \frac{1}{3}* a* h$

So, weight of small pyramid, $w = k*(\frac{1}{3}* a* h)=400$

<u>Since, base and height of small pyramid and original pyramid are in proportion.</u>

So, $\frac{a}{A} = (\frac{h}{5}) ^{2}$

$a = (\frac{h}{5} )^{2}A$

Substituting value of a in equation $k*(\frac{1}{3}* a* h)=400$

So, $k*(\frac{1}{3}* \frac{h^{2} }{25}A * h)=400$

$(k*\frac{1}{3}* A*5)*\frac{1}{5} *\frac{h^{2} }{25} * h)=400$

Since, $(k *\frac{1}{3} *A* 5) = 800$ , substitute in above equation.

So, $800*\frac{h^{3} }{125}=400\\\\h^{3}=\frac{125}{2}\\\\h=\sqrt[\frac{1}{3} ]{62.5}\\\\h=3.97 feet$

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