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4 years ago

How do you convert set notation to interval notation?

Mathematics

1 answer:

Semenov [28]4 years ago

6 0

Answer:

Imagine we are to express [ a , b ] in set notation

A = [ a , b ] , then A = { x ∈ R / a ≤ x ≤ b }

In this notation we define the characteristics of all x belonging to this set A ....x must be greater or equal to a and simultaneously smaller or equal to b...

Interval notation is other way to say the same but assuming that means the extreme a is IN the interval and means extreme a is not.

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Use the discriminant to describe the roots of each equation. Then select the best description.

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The answer is C though.
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Find the perimeter of a rectangle with a length of 12 feet and a width of 6 feet.

jeka57 [31]

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4 years ago

What is the slope that passes through the lines (2.1,-4.2) and (2.5,-5)

Usimov [2.4K]

slope = (y2-y1)/(x2-x1)

= (-5--4.2) /(2.5 -2.1)

(-5+4.2)/(2.5-2.1)

=-.8/.4

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3 years ago

Need help with linear approximation

Dmitrij [34]

Consider a function f(x), the linear approximation L(x) of f(x) is given by

$L(x)=f(a)+(x-a)f'(a)$

Given the quantity: $\frac{1}{203}$

We approximate the quantity using the function $f(x)= \frac{1}{x}$ , where x = 203.

We choose a = 200, thus the linear approximation is given as follows:

$L(203)=f(200)+(203-200)f'(200) \\ \\ = \frac{1}{200} - \frac{3}{200^2} = \frac{200-3}{200^2} = \frac{197}{40,000} =0.004925$

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3 years ago

A pyramid-shaped vat has square cross-section and stands on its tip. The dimensions at the top are 2 m × 2 m, and the depth is 5

Strike441 [17]

Answer:

the water level increases at a rate of 1.718 m/min when the depth is 4 m

Step-by-step explanation:

the volume of the pyramid is

V = (1/3)(height)(area of base) = 1/3 H*L²

For the diagonal in the pyramid

tg Ф = Side Length/ Height = L / H = x / h

where h= depth of water , x= side of the corresponding cross section

therefore x= L *h/H

the volume of the water is

v= 1/3 h*x² = 1/3 (L/H)² h³

in terms of time

v = Q*t

then

Q*t = 1/3 (L/H)² h³

h³ = 3*(H/L)² *Q *t

h = ∛(3*(H/L)² *Q *t) = ∛(3*(H/L)² *Q) *∛t = k* ∛t , where k=∛(3*(H/L)² *Q)

h = k* ∛t

then the rate of increase in depth is dh/dt

dh/dt = 1/3*k* t^(-2/3)

since

t = (h/k)³

dh/dt = 1/3*k* t^(-2/3) = 1/3*k* (h/k)³ ^(-2/3) = 1/3*k* (h/k)^(-2) = 1/3 k³ / h²

= 1/3 (3*(H/L)²*Q) / h² = (H/L)²*Q /h²

dh/dt= [H/(h*L)]²*Q

replacing values, when h=4m

dh/dt= [H/(h*L)]²*Q = [5m/(4m*2m)]² * (3m³/min)= 1.718 m/min

8 0

4 years ago