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3 years ago

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Factorising Quadratics

2 answers:

umka21 [38]3 years ago

5 0

Quadratic formula where:
$ax ^{2} + bx + c$
$\frac{ - b + { \sqrt{b^{2} - 4ac } }}{2a}$
$\frac{ - b - \sqrt{b { }^{2} - 4ac } }{2a}$
do both do get the roots of the equation

Ulleksa [173]3 years ago

3 0

Answer:

Step-by-step explanation:

A short answer is that you cannot factor this the way you are trying to do it. What to do to go on is a question. You could use the quadratic formula to get some sort of answer but if factoring is your only option, forget about it.

Here's how you can check.

a = 12
b = 1
c = -15

There is a part of the quadratic solution is the sqrt(b^2 - 4*a*c). In order to be factorable, this sqrt must result in a perfect square.

sqrt(1 - 4(12)(-15) ) = sqrt( 1 - - 720) = sqrt(721)

This is anything but a perfect square. (721 does factor, however, into 7 * 103). I03 is prime. So the long and short of it is, there's no way it will factor.

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the weightbof the rectangle is 80 grams the weight of the triangle is double the weight of the circle determine the weight if th

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The weight of the circle that is double the weight of the triangle is 20 grams.

<h3>How to find weight ?</h3>

The weight of the rectangle is 80 grams.

The weight of the triangle is double the weight of the weight of the circle.

A rectangle is made up of 2 triangles. Therefore,

weight of each triangle = 80 / 2 = 40 grams

Since, the weight of the triangle is double the weight of the circle, the weight of the circle can be calculated as follows:

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2 years ago

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3 years ago

If you subtract 14 from twice a number, the result is 6. What is the number?

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3 years ago

A toy company is considering a cube or sphere-shaped container for packaging a new product. The height of the cube would equal t

Darya [45]

Answer:

Packaging with cube will be more efficient.

Step-by-step explanation:

Given:

A cube and a sphere where the diameter of the sphere is equal to the height of the cube.

Let the height of the cube "x"

Radius of the sphere = $(\frac{x}{2})$

Formula to be used:

Surface area of the cube = $6x^2$ and Surface area of the sphere = $4\pi (r)^2$

Volume of the cube = $x^3$ and Volume of the sphere = $\frac{4\pi r^3}{3}$

We have to compare the ratio of SA and Volumes.

Ratio of SA : Ratio of their volumes :

⇒ $\frac{SA\ of\ cube\ (S_1)}{SA\ of\ sphere\ (S_2)}$ ⇒ $\frac{Volume \ of \ cube\ (V_1)}{Volume\ of\ sphere\ (V_2)}$

⇒ $\frac{6x^2}{4\pi (\frac{x}{2})^2}$ ⇒ $\frac{x^3}{\frac{4 \pi r^3}{3} }$

⇒ $\frac{6x^2}{4\pi (\frac{x^2}{4})}$ ⇒ $\frac{x^3}{\frac{4 \pi (\frac{x}{2})^3}{3} }$

⇒ $\frac{6x^2}{\pi x^2}$ ⇒ $\frac{x^3}{\frac{4 \pi (\frac{x^3}{8})}{3} }$

⇒ $\frac{6}{\pi}$ ⇒ $\frac{6}{\pi}$

⇒ approx $2$ ⇒ approx $2$

⇒ $S_1=2S_2$ ⇒ $V_1=2V_2$

Packaging of the toy with the cube will be more efficient as it has more volume comparatively.

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2 years ago