All categories

2 years ago

H(n)=-2n(2)+4; Find h(4)

Mathematics

1 answer:

anygoal [31]2 years ago

8 0

Answer:

h(4) = –12

Step-by-step explanation:

⇒ What the question is asking is that when the function h(n) = –2n(2) + 4 is h(4), what will the function repond to when solving for h(4)? So, solve for the function h(4):

h(4) = –2n(2) + 4

⇒ Since n was replaced with 4 in the function h(4), substitute any n for 4 into the function:

h(4) = –2(4)(2) + 4

⇒ Simplify:

h(4) = –16 + 4

⇒ Solve:

h(4) = –12

<u>Answer:</u> h(4) = –12

<em>I hope you understand and that this helps with your question! </em>:)

You might be interested in

Could u answer me for

Licemer1 [7]

Peers if not of similar circumstances may be biased or not understand the situation as well.

4 0

1 year ago

A school choir has 36 members. This year, there are three times as many returning members as new members. How many new members a

PtichkaEL [24]

Answer:its 12

Step-by-step explanation:

because there are 36 now and they have 3 times more each year so it is 12 + 12 + 12 = 36

8 0

2 years ago

Read 2 more answers

BRAINLIEST ANSWER!! The product of 3, and a number increased by seven, is -36. Translate the equation, then solve.

EleoNora [17]

Product is multiplication.

Let the number = x

3 *( X+7) = -36

Use distributive property:

3x +21 = -36

Subtract 21 from each side:

3x = -57

Divide both sides by 3:

x = -57 /3

x = -19

Check: 3 * (-19 +7) = 3 * -12 = -36

The number is -19

4 0

2 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

Factor each expression 30+10x-40y

Sergio039 [100]

Hello!

To factor you find the biggest number they can be divided by

The biggest number that all the number can be divided by is 10

You put that outside parenthesis

10()

Divide the expression by 10

3 + x - 4y

Put the two together

10(3 + x + 4y)

Hope this helps!

4 0

2 years ago