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kherson [118]
3 years ago
15

In triangle ABC, angle A is equal to angle C, BA=x+20,CA=4x-30, and BC=3x+14. What is the length of BC?​

Mathematics
1 answer:
Anna [14]3 years ago
3 0

Answer:

BC = 23

Step-by-step explanation:

Since ∠ A = ∠ C then the triangle is isosceles with AB = BC, that is

3x + 14 = x + 20 ( subtract x from both sides )

2x + 14 = 20 ( subtract 14 from both sides )

2x = 6 ( divide both sides by 2 )

x = 3

Thus

BC = 3x + 14 = 3(3) + 14 = 9 + 14 = 23

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Translate and simpify: subtract 18 from -11
yawa3891 [41]

Answer:

should be -29

Step-by-step explanation:

If you subtract a positive from a negative number you basically add them both and put a - onto the number

5 0
3 years ago
A football is thrown from the top of the stands, 50 feet above the ground at an initial velocity of 62 ft/sec and at an angle of
Anvisha [2.4K]

a. i. The parametric equation for the horizontal movement is x = 43.84t

ii. The parametric equation for the vertical movement is y = 50 + 43.84t

b. the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

<h2>a. Parametric equations</h2>

A parametric equation is an equation that defines a set of quantities a functions of one or more independent variables called parameters.

<h3>i. Parametric equation for the horizontal movement</h3>

The parametric equation for the horizontal movement is x = 43.84t

Since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the horizontal component of the velocity is v' = vcosФ.

So, the horizontal distance the football moves in time, t is x = vcosФt

= vtcosФ

= 62tcos45°

= 62t × 0.7071

= 43.84t

So, the parametric equation for the horizontal movement is x = 43.84t

<h3>ii Parametric equation for the vertical movement</h3>

The parametric equation for the vertical movement is y = 50 + 43.84t

Also, since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the vertical component of the velocity is v" = vsinФ.

Since the football is initially at a height of h = 50 feet, the vertical distance the football moves in time, t relative to the ground is y = 50 + vsinФt

= 50 + vtcosФ

= 50 + 62tsin45°

= 50 + 62t × 0.7071

= 50 + 43.84t

<h3>b. Location of football at maximum height relative to starting point</h3>

The location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Since the football reaches maximum height at t = 1.37 s

The x coordinate of its location at maximum height is gotten by substituting t = 1.37 into x = 48.84t

So, x = 43.84t

x = 43.84 × 1.37

x = 60.0608

x ≅ 60.1 ft

The y coordinate of the football's location at maximum height relative to the ground is y = 50 + 48.84t

The y coordinate of the football's location at maximum height relative to the starting point is y - 50 = 48.84t

So,  y - 50 = 48.84t

y - 50 = 43.84 × 1.37

y - 50 = 60.0608

y - 50 ≅ 60.1 ft

So, the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Learn ore about parametric equations here:

brainly.com/question/8674159

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2 years ago
If middle school and senior school are winning football games in the ratio 3:9 . If the Middle school won 17 games, calculate ho
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The Middle school won 17 games so the senior school won 51 games

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3 years ago
How do you write 10 to the power of 3 in word form
AysviL [449]

Answer:

I believe that it would be 10^3. (10x10x10)

Sorry if I'm wrong I'm just a little confused with the question you asked :/

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A homeowner has 20 feet of fencing material to enclose a rectangular area for his pets. the rectangular area is adjacent to a ho
VikaD [51]
5X10=50 sq ft area
Around the 3 sides= 5+10+5=20
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3 years ago
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