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alina1380 [7]
3 years ago
11

The area of the conference table in Mr. Nathan’s office must be no more than 175 ft2. If the length of the table is 18 ft more t

han the width, x, which interval can be the possible widths?
Mathematics
1 answer:
melamori03 [73]3 years ago
4 0

Answer:

0 ft < w ≤ 7 ft

Step-by-step explanation:

A=wl,  where w, l >0

A≤175 ft²

l= w+18

w(w+18)≤175

w²+18w-175≤0

zeros of are -25 and 7

(w + 25)*(w - 7) ≤ 0

w+25≤0, w-7≥0 ⇒ w ⊂ ∅

w+25≥0, w-7≤0 ⇒ -25 ≤ w ≤ 7,

as w>0, we have answer: 0 < w ≤ 7

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A person who is 3 m tall casts a shadow that is 6 m long. At the same time, a building
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13.5 meters

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A 10-foot ladder is to be placed Against the side of the building. The base of the latter must be placed in an angle of 72° With
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Answer:

Distance of foot of ladder from building: <em>37.2 inches </em>

Distance of top of ladder from building's base: <em>114 inches </em>

Step-by-step explanation:

Please refer to the figure attached in the answer area.

A right angled triangle \triangle ABC is formed by the ladder with the building where hypotenuse is the length of ladder.

Hypotenuse, <em>AC </em>= <em>10 foot </em>

Also, we are given that angle made by the base of ladder with the ground is 72^\circ.

We have to find <em>AB</em> and <em>BC</em>.

\angle BAC = 72^\circ

Using trigonometric functions:

cos \theta= \dfrac{Base}{Hypotenuse}\\cos 72^\circ= \dfrac{AB}{AC}\\\Rightarrow 0.309 = \dfrac{AB}{AC}\\\Rightarrow AB = 10 \times 0.309\\$\approx$ 3.1 foot\\$\approx$ 37.2 inches

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Distance of foot of ladder from building: <em>37.2 inches </em>

Distance of top of ladder from building's base: <em>114 inches </em>

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