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3 years ago

In the new logo, what fraction of the parallelogram is shaded?

Mathematics

1 answer:

Harman [31]3 years ago

7 0

D. 1/4 because imagine splitting that parallelogram in half, and you will see on the first half that half of that half is shaded. So that part is 1/2 and if you add the second half which isn't shaded at all, that's 0/2 so do 1/2 plus 0/2 and you will get 1/4

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I have fewer than 25 i have more then 21

Black_prince [1.1K]

You could have anywhere from 22 to 24 acorns

7 0

2 years ago

Read 2 more answers

If c(x)=4x-2 and d(x)=x^2+5 , what is (c o d) (x)

Nookie1986 [14]

Answer:

4x² + 18

Step-by-step explanation:

note that (c ○ d )(x) = c(d(x))

To evaluate substitute x = d(x) into c(x), that is

c(d(x))

= c(x² + 5)

= 4(x² + 5) - 2

= 4x² + 20 - 2

= 4x² + 18

3 0

2 years ago

To compute a student's Grade Point Average (GPA) for a term, the student's grades for each course are weighted by the number of

postnew [5]

Answer:

The student's GPA for that term is 3.07

Step-by-step explanation:

So this is a problem where we need to compute a weighted arithmetic mean.

To compute the weighted arithmetic mean of a set, we need to multiply each value of the set by their respective weight, add them, and then divide by the sum of the weights.

I will write the set in a {V,W} way, where V is the value(the grade) and W is the weight of V.

So your set G will be

G = {{3.8, 5}, {1.8,3}, {3.1,5}, {3.1,5}}.

Multiplying each value by it's respective weigth and then adding, we have:

3.8*5 + 1.8*3 + 3.1*5 + 3.1*5 = 19 + 5.4 + 15.5 + 15.5 = 55.4

The sum of the weigths is 5 + 3 + 5 + 5 = 18

So the student's GPA for that term, rounded to two decimal places, is: 55.4/18 = 3.07

4 0

2 years ago

A relation from a set X to a set Y is called a function if each element of X is related to exactly one element in Y. That is, gi

defon

Answer:

An ordered pair is a set of inputs and outputs and represents a relationship between the two values. A relation is a set of inputs and outputs, and a function is a relation with one output for each input.

here is a exstended explantion:

A relation from a set X to a set Y is called a function if each element of X is related to exactly one element in Y. That is, given an element x in X, there is only one element in Y that x is related to.

For example, consider the following sets X and Y. I'll give you a relation between them that is not a function, and one that is.

X = { 1, 2, 3 }

Y = { a , b , c, d }

Relation from X to Y (i.e., in XxY ) : { (1,a) , (2, b) , (2, c) , (3, d) }

This relation is not a function from X to Y because the element 2 in X is related to two different elements, b and c. (Note, if you transpose the ordered pairs you <i>would</i> have a function from Y to X - can you see WHY?)

Relation from X to Y that is a function: { (1,d) , (2,d) , (3, a) }

This is a function since each element from X is related to only one element in Y. Note that it is okay for two different elements in X to be related to the same element in Y. It's still a function, it's just not a one-to-one function.

8 0

3 years ago

What kind of problem do we need to use “indefinite integral” to solve? Use a real life example to explain.

anastassius [24]

Answer:

Indefinite integration acts as a tool to solve many physical problems.

There are many type of problems that require an indefinite integral to solve.

Basically indefinite integration is required when we deal with quantities that vary spatially or temporally.

As an example consider the following example:

Suppose that we need to calculate the total force on a object placed in a non- uniform field.

As an example let us consider a rod of length L that posses an charge 'q' per meter length and suppose that we place it in a non uniform electric field which is given by

$E(x)=\frac{E_{o}}{e^{kx}}$

Now in order to find the total force on the rod we cannot use the similar procedure as we can see that the force on the rod varies with the position of the rod.

But if w consider an element 'dx' of the rod at a distance 'x' from the origin the force on this element will be given by

$dF=E(x)\times qdx\\\\dF=\frac{qE_{o}}{e^{kx}}dx$

Now to find the whole force on the rod we need to sum this quantity over the whole length of the rod requiring integration, as shown

$\int dF=\int \frac{qE_{o}}{e^{kx}}dx$

Similarly there are numerous problems considering motion of particles that require applications of indefinite integration.

4 0

3 years ago