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3 years ago

Can someone help me

Mathematics

1 answer:

algol [13]3 years ago

7 0

Answer:

A)310.18g/mol

B)84.9947g/mol

C)108.01 g/mol

d)159.7g /mole(Approx)

e)Coc12 g/mol

f) A1(OH)3 g/mol

g)83.9949 g/mol

h)197.9 g/mol

i)18.01529 g/mol

Step-by-step explanation:

Hope u happy with my answer...

please mark me as brainlest

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Solve for u<br> 3(2u+5)=33<br> Simplify your answer as much as possible.

Pie

6u+15=33
6u=33-15
6u=18
u=3

U = 3 Is the answer!!!
HOPE THIS HELP!! :D

4 0

3 years ago

Help please I’ll mark you brainliest

Vikki [24]

Wouldn’t it be the 7th root of 3? It’s a 30 60 90 triangle which means x=7 on that side so x on the bottom is the x root of three?

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3 years ago

What is the diameter of a cone with height 8 m and volume 150 m3 ? (1 point 7.5 m 15 m?

Hitman42 [59]

This is the concept of volume of solid materials, we are required to find the diameter of cone with height 8 and volume 150 m^3.
volume of cone is given by;V=1/3 (pi*r^2*h)
making r^2 the subject we get;
V/(pi*h)=r^2
inserting the values in our formula we get:
150/(pi*8)=r^2
r^2=5.97
thus;
r=sqrt(5.97)=2.44
But ;
diameter=2*radius
thus
diameter=2.44*2
=4.88 m

3 0

3 years ago

Can someone pls help for brainlest

ch4aika [34]

Answer:
455 square kilometers
step-by-step explanation:
a = l • w
24 • 24
576
because there is a space where it isn’t shaded we have to find the area of that as well
11 • 11
121
now just subtract
576 - 121
455

5 0

3 years ago

Read 2 more answers

You operate a gaming Web site, www.mudbeast.net, where users must pay a small fee to log on. When you charged $3 the demand was

Doss [256]

Answer:

A) The linear relation between price and demand is:

$d=-550x+2750$

The revenue R is:

$R=-550x^2+2750x$

B) The profit functionP is:

$P=-550x^2+2750x-30$

C) The largest monthly profit is obtained with a log-on fee of $2.5 per month. This corresponds to a profit of $3407.5.

Step-by-step explanation:

We have a site where the number of log-ons depends on our monthly fee. A linear relation is established between the price (log-on fee) and the number of log-ons.

We have two points for this linear relationship:

At price x=3, the demand is d=1100.
At price x=2.5, the demand is d=1375.

We will model the relation:

$d=mx+b$

We can calculate the slope m as:

$m=\dfrac{\Delta d}{\Delta x}=\dfrac{d_2-d_1}{x_2-x_1}=\dfrac{1375-1100}{2.5-3}\\\\\\m=\dfrac{275}{-0.5}=-550$

Then, replacing one point in the linear equation, we can calculate the intercept b:

$d_1=mx_1+b\\\\1100=(-550)\cdot 3+b\\\\1100=-1650+b\\\\b=1100+1650=2750$

Then, the linear relation between demand and price is:

$d=-550x+2750$

The revenue R can be expressed as the multiplication of the price and the demand:

$R=x\cdot d=x(-550x+2750)=-550x^2+2750x$

If we have a fixed cost of $30 per month, the profit P is:

$P=R-FC=-550x^2+2750x-30$

We can maximize the profit by deriving the profit function and making it equal to zero.

$\dfrac{dP}{dx}=0\\\\\\\dfrac{dP}{dx}=-550(2x)+2750(1)=0\\\\\\-1100x+2750=0\\\\x=\dfrac{2750}{1100}=2.5$

This corresponds to a profit of:

$P(2.5)=-550(2.5)^2+2750(2.5)-30\\\\P(2.5)=-550\cdot 6.25+6875-30\\\\P(2.5)=-3437.5+6875-30\\\\P(2.5)=3407.5$

5 0

3 years ago

Can someone help me ​

Can someone help me