What are the domain and range of the function f(x) = 4?

If a triangle has sides 4, 6 and 10 and a 2nd similar triangle has a perimeter of 60, find the 2nd largest side of the 2nd trian

algol13

The first triangle has a perimeter of 4+6+10 = 20.

The 2nd longest side is 6/20 = 3/10 of the perimeter. In the larger triangle, the 2nd longest side will have length (3/10)*60 = 18.

_____
These side lengths mean the "triangle" is a line segment. That is, it has zero area.

8 0

3 years ago

A cylinder has a volume of 550 cubic feet. If the height of the cylinder is 15 feet, what is the radius of the cylinder? Use 3.1

Yuki888 [10]

Answer:

r = 3.42

Step-by-step explanation:

the equation for volume is V = pi r^2 h

550 = 3.14 r^2 15

36 2/3 = 3.14 r ^2

11.6772824 = r^2

3.41720389 = r

r = 3.42

3 0

3 years ago

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PLEASE HURRY!! What is the rule for the function that is graphed?

il63 [147K]

Pretty sure it’s C. hope this helps

8 0

2 years ago

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Someone please help!! I don't understand this

OLga [1]

Answer:

b because y=x and b is in the specified area of within and i took the test a week ago.

Step-by-step explanation:

5 0

3 years ago

Please help me!! 100 points!!

Lyrx [107]

Part A.

We could take, for example:

$\begin{cases}x\geq-3\\y\geq3\end{cases}$

For x ≥ -3 we have vertical line (x-intercept = -3) and shaded region on the right side of the line.
For y ≥ 3 we have horizontal line (y-intercept = 3) and shaded region above that line.

Part B.

Simply, substitute x and y coordinates of D and E <span>to the system of inequalities from part A, and check if its is true, what we get.
</span>
$D=(-2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}-2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}$

so point D is a solution,

$E=(2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}$

and point E is also a solution to <span>our system of inequalities.
</span>
Part C.

There are two ways we could do that. First is the same as in part B. We substitute x and y coordinates of each school and check if inequality is true or false:

$A=(-5,5)\\\\y\ \textless \ 3x-3\quad\implies\quad5\ \textless \ 3\cdot(-5)-3\quad\implies\quad5\ \textless \ -18\quad\text{False}\\\\\\
B=(-4,-2)\\\\y\ \textless \ 3x-3\quad\implies\quad-2\ \textless \ 3\cdot(-4)-3\quad\implies\quad-2\ \textless \ -15\quad\text{False}\\\\\\
C=(2,1)\\\\y\ \textless \ 3x-3\quad\implies\quad1\ \textless \ 3\cdot2-3\quad\implies\quad1\ \textless \ 3\quad\text{True}\\\\\\
D=(-2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot(-2)-3\quad\implies\quad4\ \textless \ -9\quad\text{False}\\\\\\
E=(2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot2-3\quad\implies\quad4\ \textless \ 3\quad\text{False}\\\\\\$

$F=(3,-4)\\\\y\ \textless \ 3x-3\quad\implies\quad-4\ \textless \ 3\cdot3-3\quad\implies\quad-4\ \textless \ 6\quad\text{True}\\\\\\$

So Timothy can attend schools C and F.

We can also draw a graph of that inequality (pic 2).

8 0

3 years ago

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