Ask question

Ask question

All categories

2 years ago

9

If the midpoint between (18,y) and (20,-15) is (19,-5) find the value of y.

2 answers:

Kisachek [45]2 years ago

6 0

Answer:

5

Step-by-step explanation:

zlopas [31]2 years ago

5 0

Answer:

What I did was I plotted (20,-15) and (19, -5). From the endpoint to the mid point I got 1 unit to the right and 10 units up. SO I did the same but start at the midpoint and get (18,5).

You might be interested in

5x+10=115 what is the value of x

ololo11 [35]

X = 21

Hope this helped :)

6 0

2 years ago

I'd really appreciate it if anyone could help! :)

PSYCHO15rus [73]

C (third graph, the parabola); To find if it is a function or not draw a vertical line on the graph. If the vertical line could never intercept the figure more than once anywhere on the graph then it is a function.

3 0

2 years ago

Read 2 more answers

Fie A, B, C, D patru puncte necoplanare.

rodikova [14]

Answer:

Hlooo em dng like the same time as a great day of the same time as a great day of the same time as a great day may I have a good time for

8 0

2 years ago

What is the solution to the equation 4x= 48?<br> Ox=12<br> OX= 10<br> O x= 44<br> Ox=8

svetoff [14.1K]

Answer:

12

Step-by-step explanation:

48/4=12 x=12

6 0

2 years ago

Read 2 more answers

A cylindrical can without a top is made to contain 25 3 cm of liquid. What are the dimensions of the can that will minimize the

Basile [38]

Answer:

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

Step-by-step explanation:

Given that, the volume of cylindrical can with out top is 25 cm³.

Consider the height of the can be h and radius be r.

The volume of the can is V= $\pi r^2h$

According to the problem,

$\pi r^2 h=25$

$\Rightarrow h=\frac{25}{\pi r^2}$

The surface area of the base of the can is = $\pi r^2$

The metal for the bottom will cost $2.00 per cm²

The metal cost for the base is =$(2.00× $\pi r^2$ )

The lateral surface area of the can is = $2\pi rh$

The metal for the side will cost $1.25 per cm²

The metal cost for the base is =$(1.25× $2\pi rh$ )

$=\$2.5 \pi r h$

Total cost of metal is C= 2.00 $\pi r^2$ + $2.5 \pi r h$

Putting $h=\frac{25}{\pi r^2}$

$\therefore C=2\pi r^2+2.5 \pi r \times \frac{25}{\pi r^2}$

$\Rightarrow C=2\pi r^2+ \frac{62.5}{ r}$

Differentiating with respect to r

$C'=4\pi r- \frac{62.5}{ r^2}$

Again differentiating with respect to r

$C''=4\pi + \frac{125}{ r^3}$

To find the minimize cost, we set C'=0

$4\pi r- \frac{62.5}{ r^2}=0$

$\Rightarrow 4\pi r=\frac{62.5}{ r^2}$

$\Rightarrow r^3=\frac{62.5}{ 4\pi}$

⇒r=1.71

Now,

$\left C''\right|_{x=1.71}=4\pi +\frac{125}{1.71^3}>0$

When r=1.71 cm, the metal cost will be minimum.

Therefore,

$h=\frac{25}{\pi\times 1.71^2}$

⇒h=2.72 cm

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

6 0

2 years ago