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3 years ago

Solve equation using square root property X2-8X+16=1

1 answer:

7 0

X^2-8x+16=1\\ \\x^2-8x+16-1=0\\ \\x^2-8x+15=0\\ \\\Delta = b^{2}-4ac = (-8)^{2}-4*1*15=64-60=4 \\ \\x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{8- \sqrt{4}}{2}=\frac{8-2}{2}= \frac{6}{2}=3\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{8+ \sqrt{4}}{2}=\frac{8+2}{2}= \frac{10}{2}=5\\ \\Answer : x=3 \ \ or \ \ x = 5

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Simply the expression: Pls need help fast!!!!!!

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Answer:

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Step-by-step explanation:

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3 years ago

Mr. Nelson sold 14 bags of popcorn and 21 bottles of water in today. At this rate, how many more bottles of water than bags of p

e-lub [12.9K]

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He will sell 70 bags of popcorn and 105 bottles

Step-by-step explanation:

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The function f(x) = 28(1.45)x models the number of cats and dogs adopted from a local animal shelter, where x represents the num

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Answer:

Step-by-step explanation:

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4 years ago

Emma is making a scale drawing of her farm using the scale 1cm = 2.5ft. In the drawing she drew a well with a diameter of 0.5 ce

Y_Kistochka [10]

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4 years ago

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