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3 years ago

What is the product?

Mathematics

1 answer:

Phoenix [80]3 years ago

3 0

$(x+3)(2x-1) \\ =x(2x-1)+3(2x-1) \\ =2 x^{2} -x+6x-3 \\ =2 x^{2} +5x-3$

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In the electric currect flow, it is found that the resistance (measured in the units called ohms) offered by a fixed length o

Morgarella [4.7K]

Let x be the resistance of the wire with the larger diameter.
x = (0.456 x (0.01)^2)/(0.0401)^2 = 0.0284 ohms
The answer is 0.0284 ohms.

7 0

2 years ago

A new cereal will be advertised in 30-second spots during a morning TV show for children. The ad will run over the next 60 days.

Arturiano [62]

The campaign will cost $216,000

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2 years ago

Find the surface area of the right rectangular prism shown below. units²

faust18 [17]

Answer:

67 units^2

Step-by-step explanation:

The surface area of the prism is found by

SA = 2 ( lw+lh + wh)

= 2 ( 5*1.5+ 4*1.5 + 5*4)

= 2 ( 7.5+6+20)

= 2 (33.5)

= 67

5 0

3 years ago

Read 2 more answers

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

Find an equation of the circle whose diameter has endpoints (1,-4) and (-3,6).

MAVERICK [17]

Answer:

$(x+1)^2+(y-1)^2=58$