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denpristay [2]
2 years ago
5

If p is a polynomial show that lim x→ap(x)=p(a

Mathematics
1 answer:
Lostsunrise [7]2 years ago
3 0

Let p(x) be a polynomial, and suppose that a is any real number. Prove that

lim x→a p(x) = p(a) .

 

Solution. Notice that

 

2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .

 

So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x – 2.

 

Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| < 1, so −2 < x < 0. In particular |x| < 2. So

 

|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|

= 2|x|^3 + 5|x|^2 + |x| + 2

< 2(2)^3 + 5(2)^2 + (2) + 2

= 40

 

Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2 + x − 2| < ε/40 · 40 = ε.

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Given: PRST square
Zigmanuir [339]

Answer:

7a²/16

Step-by-step explanation:

Area of the triangle PTS

½ × a × a

a²/2

Length of PS:

sqrt(a² + a²)

asqrt(2)

Length of MS:

¼asqrt(2)

Triangles MCS and TPS are similar

With sides in the ratio:

¼asqrt(2) : a

sqrt(2)/4 : 1

Area of triangle SMC:

A/(a²/2) = [(sqrt(2)/4)]²

2A/a² = 1/8

A = a²/16

Area of PTMC

= a²/2 - a²/16

= 7a²/16

Step-by-step explanation:

4 0
2 years ago
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If 9 = c, then c = 9. What algebraic equality property justifies the above statement?
netineya [11]
Symmetric property is the correct term
4 0
2 years ago
119.31 is 82% of what number
madam [21]
To solve
82% of x = 119.31
rewrite
x = 119.31/0.82
solve
x <span>≈ 145.5</span>

Hope this helps :)
5 0
2 years ago
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Parallel lines s and t are cut by a transversal r. Parallel lines s and t are cut by transversal r. Clockwise from top left, the
bearhunter [10]

Answer:

The corresponding angles are:

∠ 1 and ∠ 5

∠ 2 and ∠ 6

∠ 3 and ∠ 7

∠ 4 and ∠ 8

Step-by-step explanation:

When two parallel lines are crossed by another line, the angles in matching corners are called corresponding angles and they are equal in measures

∵ Lines s and t are parallel

∴ s // t

∵ Line r intersects lines s and t

∴ Line r is the transversal

∵ Corresponding angles are formed

- Look to the attached figure

∴ ∠ 1 ≅ ∠ 5 ⇒ corresponding angles

∴ ∠ 2 ≅ ∠ 6 ⇒ corresponding angles

∴ ∠ 3 ≅ ∠ 7 ⇒ corresponding angles

∴ ∠ 4 ≅ ∠ 8 ⇒ corresponding angles

The corresponding angles are:

∠ 1 and ∠ 5

∠ 2 and ∠ 6

∠ 3 and ∠ 7

∠ 4 and ∠ 8

4 0
2 years ago
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Barbara drives between Miami, Florida, and West Palm Beach, Florida. She drives 50 mi in clear weather and then encounters a thu
NeX [460]

Distance traveled in clear weather = 50 miles

Distance traveled in thunderstorm = 15 miles

Let speed in clear weather = x

⇒ Speed in thunderstorm = x-20

Total time taken for trip = 1.5 hours

We need to determine average speed in clear weather (i.e. x) and average speed in the thunderstorm (i.e. x-20 ).

Total time taken for trip = Time taken in clear weather + Time taken in thunderstorm

⇒ Total time taken for trip = \frac{Distance covered in clear weather}{Speed in clear weather} + \frac{Distance covered in thunderstorm}{Speed in thunderstorm}

⇒ 1.5 = \frac{50}{x} + \frac{15}{x-20}

⇒ 1.5 = \frac{50(x-20)+15(x)}{(x)(x-20)}

⇒ 15*x*(x-20) = 10*[50*(x-20)+15*x]

⇒ 15x² - 300x = 500x - 10,000 + 150x

⇒ 15x² - 300x = 650x - 10,000

⇒ 15x² - 950x + 10,000 = 0

⇒ 3x² - 190x + 2,000 = 0

The above equation is in the format of ax² + bx + c = 0

To determine the roots of the equation, we will first determine 'D'

D = b² - 4ac

⇒ D = (-190)² - 4*3*2,000

⇒ D = 36,100 - 24,000

⇒ D = 12,100

Now using the D to determine the two roots of the equation

Roots are: x₁ = \frac{-b+\sqrt{D}}{2a} ; x₂ = \frac{-b-\sqrt{D}}{2a}

⇒ x₁ = \frac{-(-190)+\sqrt{12,100}}{2*3} and x₂ = \frac{-(-190)-\sqrt{12,100}}{2*3}

⇒ x₁ = \frac{190+110}{6} and x₂ = \frac{190-110}{6}

⇒ x₁ = \frac{300}{6} and x₂ = \frac{80}{6}

⇒ x₁ = 50 and x₂ = 13.33

So speed in clear weather can be 50 mph or 13.33 mph. However, we know that in thunderstorm was 20 mph less than speed in clear weather.

If speed in clear weather is 13.33 mph then speed in thunderstorm would be negative, which is not possible since speed can't be negative.

Hence, the speed in clear weather would be 50 mph, and in thunderstorm would be 20 mph less, i.e. 30 mph.

7 0
2 years ago
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