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3 years ago

MIN, median or Q2, Q1, or MAX

Mathematics

1 answer:

ikadub [295]3 years ago

3 0

Answer:

median or Q2

hope this helps

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Jeremiah is working on a model bridge. He needs to create triangular components, and he plans to use toothpicks. He finds three

icang [17]

Answer: No

Step-by-step explanation:

According to triangle inequality theorem which state that: the sum of any 2 sides of a triangle must be greater than the measure of the third side.

Now let's test the given parameters if they obey this theorem

4 inches, 5 inches and 1 inch

4 + 5 = 9

9 > 1 correct

5 + 1 = 6

6 > 4 correct

4 + 1 = 5

5 not > 5 wrong

Therefore he won't be able to create the triangular component with these toothpicks without modifying any of the lengths

3 0

4 years ago

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Given the function defined in the table below, find the average rate of change, in the simplest form, of the function over the i

vekshin1

Answer:

im stuck on that question too funny bc im in your class and caught u cheating

Step-by-step explanation:

4 0

3 years ago

I neeeeeed help really fast, please!! 0 X 0 = ?

Fiesta28 [93]

The answer is zero 0

6 0

3 years ago

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Suppose a city with population 600,000 has been growing at a rate of 3% per year. If this rate continues, find the population

andrezito [222]

Answer:

The answer should be around 1,149663 (Rounded off to the nearest whole number. Also if a population has a remainer you usually add it as 1. This is because these are people)

Step-by-step explanation:

First step involves listing the formulas being used:

600,000= population, 0.03= growth rate 22years= period

Therefore, we take the 600,000 and multiply by 1+0.03 ^22(to the power of 22 or period)

So, 600,000(1+0.03)^22

=1,149662.045 /1,149663

Hope this helps!

6 0

3 years ago

Find the solution of the given initial value problems in explicit form. Determine the interval where the solutions are defined.

natali 33 [55]

Answer:

The solution of the given initial value problems in explicit form is $y=x-x^2-2$ and the solutions are defined for all real numbers.

Step-by-step explanation:

The given differential equation is

$y'=1-2x$

It can be written as

$\frac{dy}{dx}=1-2x$

Use variable separable method to solve this differential equation.

$dy=(1-2x)dx$

Integrate both the sides.

$\int dy=\int (1-2x)dx$

$y=x-2(\frac{x^2}{2})+C$ $[\because \int x^n=\frac{x^{n+1}}{n+1}]$

$y=x-x^2+C$ ... (1)

It is given that y(1) = -2. Substitute x=1 and y=-2 to find the value of C.

$-2=1-(1)^2+C$

$-2=1-1+C$

$-2=C$

The value of C is -2. Substitute C=-2 in equation (1).

$y=x-x^2-2$

Therefore the solution of the given initial value problems in explicit form is $y=x-x^2-2$ .

The solution is quadratic function, so it is defined for all real values.

Therefore the solutions are defined for all real numbers.

4 0

3 years ago