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3 years ago

Complete the patterns

Mathematics

1 answer:

Rzqust [24]3 years ago

7 0

A. 0.02 0.2 2 20 200 2000
b. 3,400,000 , 34,000 , 340 ,3.4, 0.034
c. 85700 8,570 857 85.7 8.57 0.857
d. 444 4440 44,400 444,000 4,440,000 44,400,000
e. 0.095 9.5 950 95,000 9,500,000 950,000,000

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You invest $5000 in an account at 5.5% per year simple interest. How much will you have in the account at the beginning of the 6

NeX [460]

Answer:

The amount in the account in the beginning of the 6th year is $6375 .

Step-by-step explanation:

Formula

$Simple\ interest = \frac{Principle\times Rate\ times Time}{100}$

As given

You invest $5000 in an account at 5.5% per year simple interest.

Principle = $ 5000

Rate = 5.5 %

Time = 5 years

(As calculate amount in the account for beginning of 6th year.)

Putting all the values in the formula

$Simple\ interest = \frac{5000\times 5.5\times 5}{100}$

$Simple\ interest = 50\times 5.5\times 5$

Simple interest = $ 1375

Amount = Principle + Simple interest

Putting values in the above formula

Amount = $5000 + $1375

Amount = $ 6375

Therefore the amount in the account in the beginning of the 6th year is $ 6375 .

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3 years ago

DIscrete Math

Daniel [21]

Answer:

Step-by-step explanation:

As the statement is ‘‘if and only if’’ we need to prove two implications

$f : X \rightarrow Y$ is surjective implies there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ .
If there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ , then $f : X \rightarrow Y$ is surjective

Let us start by the first implication.

Our hypothesis is that the function $f : X \rightarrow Y$ is surjective. From this we know that for every $y\in Y$ there exist, at least, one $x\in X$ such that $y=f(x)$ .

Now, define the sets $X_y = \{x\in X: y=f(x)\}$ . Notice that the set $X_y$ is the pre-image of the element $y$ . Also, from the fact that $f$ is a function we deduce that $X_{y_1}\cap X_{y_2}=\emptyset$ , and because $f$ the sets $X_y$ are no empty.

From each set $X_y$ choose only one element $x_y$ , and notice that $f(x_y)=y$ .

So, we can define the function $h:Y\rightarrow X$ as $h(y)=x_y$ . It is no difficult to conclude that $f\circ h(y) = f(x_y)=y$ . With this we have that $f\circ h=1_Y$ , and the prove is complete.

Now, let us prove the second implication.

We have that there exists a function $h:Y\rightarrow X$ such that $f\circ h=1_Y$ .

Take an element $y\in Y$ , then $f\circ h(y)=y$ . Now, write $x=h(y)$ and notice that $x\in X$ . Also, with this we have that $f(x)=y$ .

So, for every element $y\in Y$ we have found that an element $x\in X$ (recall that $x=h(y)$ ) such that $y=f(x)$ , which is equivalent to the fact that $f$ is surjective. Therefore, the prove is complete.

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Rom4ik [11]

Answer:

divided

Step-by-step explanation:

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It should be bigger, think of a pie:
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3 years ago

Lewis needed to find the difference in lengths between the hypotenuse and the longest leg of this triangle. His work is shown be

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Answer:

The answer is first and last

Step-by-step explanation:

Edge2020

im not gonna explain cuz some other guy got it wrong and im just answering for anyone who needs it

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3 years ago

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