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2 years ago

Prove that 3(x+1)(x+7)-(2x+5)² is never positive

Mathematics

2 answers:

scZoUnD [109]2 years ago

3 0

Step-by-step explanation:

3(x + 1)(x + 7) − (2x + 5)²

3(x² + 8x + 7) − (4x² + 20x + 25)

3x² + 24x + 21 − 4x² − 20x − 25

-x² + 4x − 4

-(x² − 4x + 4)

-(x − 2)²

A squared number is never negative (provided x is a real number), so -(x − 2)² is never positive.

Reika [66]2 years ago

3 0

Answer:

-27x - 219

Step-by-step explanation:

3x+3 * 3x+21 - 6x+15

9x+6 - (6x+15 * 6x+15)

9x+6 - 36x+325

-27x - 219

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Step-by-step explanation:

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Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the

Rama09 [41]

Question:

Britney throws an object straight up into the air with an initial velocity of 27 ft/s from a platform that is 10 ft above the ground. Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the object to hit the ground?

1s 2s 3s 4s

Answer:

It takes 2 seconds for object to hit the ground

Solution:

The given equation is:

$h(t) = -16t^2 + v_0t+h_0$

Initial velocity = 27 feet/sec

$h_0 = 10\ feet$

Therefore,

$h(t) = -16t^2 +27t+10$

At the point the object hits the ground, h(t) = 0

$-16t^2 +27t+10 = 0\\\\16t^2-27t-10=0$

Solve by quadratic formula,

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=16,\:b=-27,\:c=-10:\quad \\\\t=\frac{-\left(-27\right)\pm \sqrt{\left(-27\right)^2-4\cdot \:16\left(-10\right)}}{2\cdot \:16}\\\\t = \frac{27 \pm \sqrt{1369}}{32}\\\\t = \frac{27 \pm 37}{32}\\\\We\ have\ two\ solutions\\\\t = \frac{27+37}{32}\\\\t = \frac{64}{32}\\\\t = 2\\\\And\\\\t = \frac{27-37}{32}\\\\t = -0.3125$

Ignore, negative value

Thus, it takes 2 seconds for object to hit the ground

7 0

3 years ago