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Mathematics

1 answer:

yKpoI14uk [10]3 years ago

8 0

Answer:

1. $x^\circ= 7^\circ$

2.(a)Therefore , Side LJ =side CA

(b) BC = KL

Step-by-step explanation:

we know that, vertically opposite angles are congruent .

$\therefore (x+16)^\circ= (4x-5)^\circ$

$\Leftrightarrow 4x^\circ-x^\circ= 5^\circ+16^\circ$

$\Leftrightarrow 3x^\circ= 21^\circ$

$\Leftrightarrow x^\circ= 7^\circ$

(a)SAS = side,angle , side

Given ,side AB = side KJ

and ∠CAB= ∠LJK

All Pythagorean Triples are unique.

Since side AB = side KJ

Therefore , Side LJ =side CA

It satisfies SAS postulate ThereforeΔABC≅ΔKJL

(b)

HL Theorem: If hypotenuse and one leg of of a right triangle are congruent to the hypotenuse and one leg of of other right triangle , then the triangle congruent.

Since,

All Pythagorean Triples are unique.

Then hypotenuse of ΔABC=ΔLKJ ⇔BC = KL

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A projectile is launched from ground level with a initial velocity of Vo feet per second. Neglecting air resistance, its height

antiseptic1488 [7]

The height at t seconds after launch is
s(t) = - 16t² + V₀t
where V₀ = initial launch velocity.

Part a:
When s = 192 ft, and V₀ = 112 ft/s, then
-16t² + 112t = 192
16t² - 112t + 192 = 0
t² - 7t + 12 = 0
(t - 3)(t - 4) = 0
t = 3 s, or t = 4 s

The projectile reaches a height of 192 ft at 3 s on the way up, and at 4 s on the way down.

Part b:
When the projectile reaches the ground, s = 0.
Therefore
-16t² + 112t = 0
-16t(t - 7) = 0
t = 0 or t = 7 s

When t=0, the projectile is launched.
When t = 7 s, the projectile returns to the ground.

Answer: 7 s