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3 years ago

Help me in math. I need better grads

Mathematics

2 answers:

RoseWind [281]3 years ago

8 0

I have no idea sorry

elena-s [515]3 years ago

4 0

58.6 mph............

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Which expression is equivalent to (4−3)−6?

arsen [322]

Answer:

The given expression is equal to -5

Step-by-step explanation:

Given that:

= (4−3)−6

solving bracket first as 4 -3 equals 1

= 1-6

Subtracting greater term from smaller gives the negative value

= -5

So the given expression will be equal to -5

i hope it will help you!

5 0

4 years ago

Read 2 more answers

-3ab+b2+a2 i need help plz

vfiekz [6]

Answer:

a2 - 3ab + b2

Step-by-step explanation:

just saying the 2s are squared :)

7 0

4 years ago

Ninety-eight thousand ,five (how to write in numerical)?

Anarel [89]

Answer:

98005

Step-by-step explanation:

ninety-eight thousand = 98000

five = 5

together = 98005

7 0

3 years ago

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

3 years ago

About how much would it have cost

Anton [14]

It would cost about $.50 because $8.15 rounds to $8.00, and 8/16 = 0.5

4 0

3 years ago

Read 2 more answers