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2 years ago

Convert solar days into minutes

Mathematics

1 answer:

Pepsi [2]2 years ago

6 0

Answer:

1 Solar day = 86,400 seconds (SI Base Unit)

1 day = 1,440 minutes

Step-by-step explanation:

If need anymore help just ask I'm not really good at explaining but I'll try just comment haha ☺️

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Using the equation below, find the slope and y-intercept of the equation it would create.

IRISSAK [1]

Answer: m=-1, b=-2

Step-by-step explanation:

4 0

3 years ago

e) Given the following: Let X = {1, 2, 3, 4) and a relation R on X as R= {(1,2), (2,3), (3,4)}. Find the reflexive and transitiv

Naddika [18.5K]

Answer:

The answer is $\{(1,1),(2,2),(3,3),(4,4),(1,2)(2,3)(3,4),(1,3),(1,4)\}$

Step-by-step explanation:

Remember that a reflexive relation $R\subset \mathcal{P}(X)$ , where $\mathcal{P}(X)$ is the power set of $X$ , is one which conteins the ordered pairs of the form $(a,a)$ , for $a\in X$ .

So, As the reflexive and transitive closure of $R$ (that we will denote by $\overline{R}$ ) is in particular reflexive, we must add to $R$ the elements $\{(1,1) , (2,2),(3,3),(4,4) \}$

A transitive relation $R$ is one in which if the pair $(a,b)$ and the pair $(b,c)$ are in there, then the pair $(a,c)$ must be there too.

So, to complete the relation $R$ to be reflexive and transitive we must add the pair $(1,3)$ (because $(1,2),(2,3)$ are in $R$ ), the pair $(2,4)$ , and the pair $(1,4)$ because we added the pair $(2,4)$ .

Therfore we have that $\overline{R}=\{(1,1),(2,2),(3,3),(4,4),(1,2)(2,3)(3,4),(1,3),(1,4)\}$ .

6 0

3 years ago

A Norman window is a window with a semi-circle on top of regular rectangular window. (See the picture.) What should be the dimen

Vikki [24]

Answer:

bottom side (a) = 3.36 ft

lateral side (b) = 4.68 ft

Step-by-step explanation:

We have to maximize the area of the window, subject to a constraint in the perimeter of the window.

If we defined a as the bottom side, and b as the lateral side, we have the area defined as:

$A=A_r+A_c/2=a\cdot b+\dfrac{\pi r^2}{2}=ab+\dfrac{\pi}{2}\left (\dfrac{a}{2}\right)^2=ab+\dfrac{\pi a^2}{8}$

The restriction is that the perimeter have to be 12 ft at most:

$P=(a+2b)+\dfrac{\pi a}{2}=2b+a+(\dfrac{\pi}{2}) a=2b+(1+\dfrac{\pi}{2})a=12$

We can express b in function of a as:

$2b+(1+\dfrac{\pi}{2})a=12\\\\\\2b=12-(1+\dfrac{\pi}{2})a\\\\\\b=6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a$

Then, the area become:

$A=ab+\dfrac{\pi a^2}{8}=a(6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a)+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a^2+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}-\dfrac{\pi}{8}\right)a^2\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{8}\right)a^2$