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igomit [66]
3 years ago
11

PLEASE HURRY

Mathematics
1 answer:
s2008m [1.1K]3 years ago
8 0

Answer:

1.25p

Step-by-step explanation:

p = amount raised last year

amount raised this year = p +(25% of p)

=1 p+0.25p

=p(1+0.25)

=1.25p

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18.96 x 2.03 correct to 2 significant figures equals what?
Andreas93 [3]

9514 1404 393

Answer:

  38

Step-by-step explanation:

The product of the given numbers is 38.4888. Rounding this to two significant digits gets you 38.

6 0
2 years ago
Given that the restaurant is located in a city with a population over 500,000, what is the probability that it is in the northea
xxMikexx [17]
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5 0
3 years ago
What is the equation of the line that is perpendicular to the given line and has an x-intercept of -3?
Anvisha [2.4K]

Answer:

y=-\frac{3}{2}x-\frac{9}{2}

Step-by-step explanation:

Let the equation of the perpendicular line is,

y = mx + b

where m = slope of the line

b = y-intercept

From the graph, slope of the line passing through (0, -1) and (3, 1),

m' = \frac{y_2-y_1}{x_2-x_1}

m' = \frac{1+1}{3-0}

m' = \frac{2}{3}

To get the slope (m) of this line we will use the property of perpendicular lines,

m × m' = (-1)

m × \frac{2}{3} = -1

m = -\frac{3}{2}

Equation of the perpendicular line will be,

y=-\frac{3}{2}x+b

x-intercept of the line is (-3) therefore, point on the line is (-3, 0)

0 = -\frac{3}{2}(-3)+b

b = -\frac{9}{2}=-4.5

Equation of the line will be,

y=-\frac{3}{2}x-\frac{9}{2}

3 0
3 years ago
Hi I’m bored UwU<br> So who wanna talk?
ryzh [129]
Sureeeeeeeeeeeeeeeeeeee
3 0
2 years ago
If p is a polynomial show that lim x→ap(x)=p(a
Lostsunrise [7]

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.

3 0
2 years ago
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