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3 years ago

Help!!! 30pts for whoever answers!! Please helpppp.

Mathematics

2 answers:

umka21 [38]3 years ago

8 0

A simple way to do this is to keep on adding the cost per mile the same time you add the cost per day for both until you get the same number hope this helps!<3

VARVARA [1.3K]3 years ago

4 0

Answer:

i am srry i dont know but maybe u can multiply the cost per day by cost per mile then subtract or divide

by 2

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1. When you multiply an equation by a value, are you changing the solutions of that equation? Explain. 44 POINTS!!

jolli1 [7]

Answer: Yes. When you solve your equations, you're solutions will change a little bit. It's good because it'll give you the right answer.

I hope this helps you and I hope it's right. ☺

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3 years ago

Distances in space are measured in light-years. The distance from Earth to a star is 6.8 × 1013 kilometers. What is the distance

aalyn [17]

6.8 x 10^13 = 68,000,000,000,000
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3 years ago

In the division problem, 0.08 ÷ 2 = 0.04, the_____ is 0.08

-Dominant- [34]

In the division problem, 0.08 ÷ 2 = 0.04, the dividend is 0.08.

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2 years ago

Read 2 more answers

At the first tri-city meeting, there were 8 people from town A, 7 people from town B, and

SIZIF [17.4K]

Answer:

i think the answer is town a will be 6, town b 5 town c is 4

Step-by-step explanation:

first you gather information that you already know
second you put each number with the correct town
last you subtract the probability of the numbers like 2-town a 8 which 6

3 0

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.

3 0

2 years ago