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Elis [28]
3 years ago
14

What is the initial value of the function represented by this graph?

Mathematics
1 answer:
Travka [436]3 years ago
6 0

Answer:

2

Step-by-step explanation:

I didn't know what an initial value of a graph was at first until I looked it up online and saw that it's basically the y-intercept ,so the y-intercept of the graph would be the value or number that the line is hitting in the y - axis and that would be 2 :-)

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3 0
3 years ago
David wants to measure the width of a road
alexira [117]

Answer:

<u>The width of the road =  25.359 m</u>

Step-by-step explanation:

See the attached figure which represents the problem.

The length of AB = 40 m

let the width of the road w.

Construct PC perpendicular to AB, So, the measure of angle C = 90°

∠B = 60° and ∠A = 45°

Let the length of AC = x, so, the length of BC = 40-x

At ΔBCP which is a right triangle at C

tan B = opposite/adjacent = w/(40-x)

w = (40-x) * tan B  ⇒(1)

At ΔACP which is a right triangle at C

tan A = opposite/adjacent = w/x

w = x * tan A  ⇒(2)

from (1) and (2)

(40-x) * tan B = x * tan A

40 tan B - x *tan B = x tan A

40 tan B  = x tan A + x *tan B

40 tan B  = x (tan A + tan B)

x = (40 tan B)/(tan A + tan B) = (40 tan 60)/(tan 60 + tan 45) = 25.359 m

substitute at (2) with x

w = x tan A = 25.359 tan 45 = 25.359 m

<u>So, The width of the road =  25.359 m</u>

8 0
4 years ago
In a restaurant john bought one cup of gumbo and two salads for $12.25. Greg bought 4 gumbos and 4 salads for $31 how much does
Sever21 [200]
The answer is some number
7 0
4 years ago
PLS HELP ME!!!!!!!! 99 POINTS
Amiraneli [1.4K]

Answer:

\angle LKN=\arctan1.8\sqrt{97}\approx 86.77^{\circ}

Step-by-step explanation:

Draw the second altitude MB (see attached diagram).

Quadrilateral HLMB is a rectangle, then LM = HB = 3 units.

Trapezoid KLMN is isosceles trapezoid (because KL=MN), thus

KH = BN = \dfrac{KN-HB}{2}=\dfrac{13-3}{2}=5\ units\\ \\HN=HB+BN=3+5=8\ units

Triangle LHN is right triangle, then by Pythagorean theorem,

LN^2=LH^2+HN^2\\ \\89^2=LH^2+8^2\\ \\LH^2=89^2-8^2\\ \\LH^2=(89-8)(89+8)\\ \\LH^2=81\cdot 97\\ \\LH=9\sqrt{97}\ units

Consider right triangle KLH. In this triangle,

\tan\angle LKH=\{\tan\angle LKH\}=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{LH}{KH}=\dfrac{9\sqrt{97}}{5}=1.8\sqrt{97}\ units

So,

\angle LKN=\arctan1.8\sqrt{97}\approx 86.77^{\circ}

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3 years ago
Simplify x2+4x+4/(x+2)2
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8x^2+4 = x2+4x/(x+2)2
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