Pls help me. First answer will be marked brainliest

Question 2 (Multiple Choice Worth 1 points)

Alexxx [7]

I think it is d. the last one

6 0

4 years ago

Read 2 more answers

The domain of the function is all<br> The range of the function is all

kondaur [170]

The domain of a function is the set of all the x terms of the function and the range of a function is the set of all the y terms of a function.

For example, take a look below.

The domain is the set of all x terms in each ordered pair and the

range will be the set of all the y terms in each ordered pair.

7 0

3 years ago

There are 3.79 L in every gallon how many liters of gas does a 13.6 gallon tank hold

Zinaida [17]

Answer:

51.544 L in our tank

Step-by-step explanation:

We have a 13.6 gallon tank

Our conversion factor is 3.79 L = 1 gallon. We want liters on top since that is where we want to end up

3.79 L /1 gallon is what we want to use

13.6 * 3.79L/1 gallon = 51.544 L in our tank

4 0

3 years ago

What is the solution set for this inequality

Zanzabum

Answer:

Letter c po sorry nasa daan po kasi ako kaya di po ako makakapag explain

Step-by-step explanation:

Sorry po talaga pero sigurado po ako sa sagot ko

4 0

3 years ago

Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0

4 years ago