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3 years ago

5

What is the sum of the first 10,000 positive integers?

1 answer:

blagie [28]3 years ago

7 0

The sum of the first n positive integers can be easily found using Gauss' method.

$\sum\limits_{i=1}^n i=\dfrac{n(n+1)}{2}\\\\
\sum\limits_{i=1}^{10,000} i=\dfrac{10,000(10,000+1)}{2}\\
\sum\limits_{i=1}^{10,000} i=\dfrac{10,000 \cdot10,001}{2}\\
\sum\limits_{i=1}^{10,000} i=50,005$

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Answer: It would be 360 gallons of water per 10 mins because

Step-by-step explanation:

You multiply 36×10 and you get 360 gallons

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3 years ago

A business owner noted the features of the 100 cars parked at the business. Here are the results:

AlexFokin [52]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability is $P(4 \ doors | sunroof ) = 0.40$

Step-by-step explanation:

From the question we are told that

The number of cars packed is $n = 100$

The objective of this solution is to find the probability that the car which has a sunroof also has four doors

This probability is mathematically represented as

$P(4 \ doors | sunroof )$

Baye's Rule define this probability mathematically as

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From the diagram

$P(4 \ doors \ n \ sunroof ) = 20$

and $P(sunroof) = 50$

So

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4 years ago

Bottles of sparkling water usually cost $1.69 each. This week they are in sale for 4 for $5. You bought one last week and one th

Phantasy [73]

Answer:

The cost of 1 sparkling bottle in this week is less than last week, so we pay less this week.

We pay $0.44 less this week.

Step-by-step explanation:

Given,

Cost of 1 sparkling bottle last week = $1.69

This week cost of 4 bottles is $5.

So we have to find the cost of 1 bottle.

For this we use the unitary method, we get the cost of 1 bottle;

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The cost of 1 sparkling bottle in this week is less than last week, so we pay less this week.

For finding how much we pay less this week, we have to subtract cost of 1 sparkling bottle this week from cost of 1 sparkling bottle last week.

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4 years ago

Geometry Homework Help Please?

kolbaska11 [484]

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3 years ago

Let $f(x) = 2x^2 + 3x - 9,$ $g(x) = 5x + 11,$ and $h(x) = -3x^2 + 1.$ Find $f(x) - g(x) + h(x).$

Viefleur [7K]

QUESTION 1

Given that:

$f(x)=2x^2+3x-9$ ,

$g(x)=5x+11$ ,

and

$h(x)=-3x^2+1$

Then;

$f(x)-g(x)+h(x)=2x^2+3x-9-(5x+11)+(-3x^2+1)$

$f(x)-g(x)+h(x)=2x^2+3x-9-5x-11-3x^2+1$

Group similar terms;

$f(x)-g(x)+h(x)=2x^2-3x^2+3x-5x-11-9+1$

Simplify;

$f(x)-g(x)+h(x)=-x^2-2x-19$

QUESTION 2

Given that;

$f(x)=4x-7$ .

$g(x)=(x+1)^2$

and

$s(x)=f(x)+g(x)$

Substitute the functions;

$s(x)=4x-7+(x+1)^2$

Substitute x=3

$s(3)=4(3)-7+(3+1)^2$

$s(3)=12-7+(4)^2$

$s(3)=5+16$

$s(3)=21$

QUESTION 3

Given:

$f(x)=3x+2$

$g(x)=x^2-5x-1$

$f(g(x))=f(x^2-5x-1)$

This implies that;

$f(g(x))=3(x^2-5x-1)+2$

Expand the parenthesis;

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$f(g(x))=3x^2-15x-1$

QUESTION 4

The given function is;

$f(x)=3(x-6)^2+1$

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$y=3(x-6)^2+1$

$\Rightarrow y-1=3(x-6)^2$

$\Rightarrow \frac{y-1}{3}=(x-6)^2$

$\Rightarrow \sqrt{\frac{y-1}{3}}=x-6$

$\Rightarrow x=6+\sqrt{\frac{y-1}{3}}$

The range is:

$\frac{y-1}{3}\ge0$

$y-1\ge0$

$y\ge1$

The interval notation is;

$[1,+\infty)$

6 0

4 years ago