By what percent will a fraction decrease if its numerator is increased by 20% and its denominator is increased by 50%?

PLEASE HELP SOLVE! FIRST TO SOLVE RIGHT WILL GET BRAINIEST

IrinaVladis [17]

Answer:

40π/3 cm^2

Step-by-step explanation:

The centerline of the shaded region has a radius of 3 +4/2 = 5 cm. Its length is 1/3 of a circle with that radius, so is ...

length of centerline = (1/3)(2π·5 cm) = (10/3)π cm

The shaded region is 4 cm wide, so the area is the product of that width and the centerline length:

(4 cm)(10/3 π cm) = 40π/3 cm^2

3 0

3 years ago

The coefficient of 8 · 2N is ___.

MrMuchimi

The coefficient would be 2

3 0

3 years ago

Mary is making some shirts for her school’s drama department. De fabric store has three and one sixths yards of the fabric she w

Keith_Richards [23]

6 and one third. Hope this helps!

4 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

3 years ago

Will give brainliest !! Helpppp please!

Drupady [299]

The answer is A your answer

6 0

3 years ago