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mr_godi [17]
4 years ago
10

If the sum of the legs of a right triangle is 49 inches and the hypotenuse is 41 inches, find the length of the other two sides

Mathematics
1 answer:
german4 years ago
4 0
The length of one leg is x inches.
The sum of the legs is 49 inches, so the length of the other leg is 49-x inches.
The length of the hypotenuse is 41 inches.

Use the Pythagorean theorem:
(\hbox{one leg})^2 + (\hbox{the other leg})^2=(\hbox{hypotenuse})^2 \\ x^2+(49-x)^2=41^2 \\
x^2+2401-98x+x^2=1681 \\
2x^2-98x+2401=1681 \ \ \ |-1681 \\
2x^2-98x+720=0 \\ \\
a=2 \\ b=-98 \\ c=720 \\ b^2-4ac=(-98)^2-4 \times 2 \times 720=9604-5760=3844 \\ \\
x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-98) \pm \sqrt{3844}}{2 \times 2}=\frac{98 \pm 62}{4} \\
x=\frac{98-62}{4} \ \lor \ x=\frac{98+62}{4} \\
x=\frac{36}{4} \ \lor \ x=\frac{160}{4} \\
x=9 \ \lor \ x=40


49-x=49-9 \ \lor \ 49-x=49-40 \\
49-x=40 \ \lor \ 49-x=9

The lengths of the legs are 9 inches and 40 inches.
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fiasKO [112]

Answer:

see below

Step-by-step explanation:

1. 10/16=5/6

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6.5/10=1/2

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14.28/35=4/5

8 0
3 years ago
21 - 3 + 8 or -14 + 31 - 6
sergejj [24]

Answer:21-3+8=26

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Step-by-step explanation:

3 0
4 years ago
What is the circumference of a 9 m circle?
zhenek [66]
From what I know its 56.52m
5 0
3 years ago
A \greenD{7\,\text{cm} \times 5\,\text{cm}}7cm×5cmstart color #1fab54, 7, start text, c, m, end text, times, 5, start text, c, m
erma4kov [3.2K]

Answer:

The area of the shaded region is 148.04 cm².

Step-by-step explanation:

It is provided that a 7 cm × 5 cm rectangle is inside a circle with radius 6 cm.

The sides of the rectangle are:

l = 7 cm

b = 6 cm.

The radius of the circle is, r = 6 cm.

Compute the area of the shaded region as follows:

Area of the shaded region = Area of rectangle - Area of circle

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Thus, the area of the shaded region is 148.04 cm².

8 0
3 years ago
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równanie zmiany prędkość autokaru poruszającego się po prostym odcinku szosy i rozpoczynającego hamowanie od szybkości 20m/s ma
Rasek [7]

Answer:

s = 22.5 m

Step-by-step explanation:

the equation for the speed change of a coach moving along a straight section of the road and starting braking at a speed of 20 m / s has the form v (t) = 25-5t. Using integral calculus, determine the coach's braking distance.

v (t) = 25 - 5 t

at t = 0 , v = 20 m/s

Let the distance is s.

s =\int v(t) dt\\\\s =\int (25 - 5t)dt\\\\s= 25 t - 2.5 t^2 \\

Let at t = t, the v = 20

So,

20 = 25 - 5 t

t = 1 s

So, s = 25 x 1 - 2.5 x 1 = 22.5 m

3 0
3 years ago
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