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3 years ago

4. If the balloon is 800 meters high, what is the farthest distance that you would be able to see?

Mathematics

1 answer:

Xelga [282]3 years ago

6 0

yes dgsjshjegsgshsgehajgsjsjdgdjshshshjsjsjsjsjs

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Write as a sum, difference or multiple of logarithms. In(8x) = [ ? ]8+ [ ]x

deff fn [24]

Answer:

In(8x)=[In]8 + [In]x

Step-by-step explanation:

By product rule:

$In(8x)=[In]8 + [In]x$

8 0

3 years ago

The length of a rectangle is 8 feet more than its width. If the width is increased by 4 feet and the length is decreased by 5 fe

alexgriva [62]

Let the width be x.

Length is 8 feet more than width. Length = x + 8

Area = x(x + 8)

width increased by 4, that is, (x + 4)
Length decreased by 5, (x + 8 - 5) = (x + 3)

Area = (x + 4)(x +3)

Area remains the same

x(x + 8) = (x+4)(x +3)

x² + 8x = x(x +3) + 4(x +3)

x² + 8x = x² +3x + 4x +12

x² + 8x = x² +7x +12 Eliminate x² from both sides

8x = 7x + 12

8x - 7x = 12

x = 12

Dimensions of original rectangle : x, x + 8

12, 12 +8 = 12, 20

Original rectangle is 20 feet by 12 feet

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3 years ago

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What is the simpest form of 69 over 30?

avanturin [10]

2.3 in decimal form

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3 years ago

The Spotlight on Teaching and Learning at the beginning of this chapter discusses the emphasis in the Common Core State Standard

ivanzaharov [21]

I don't know this question

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3 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