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Usimov [2.4K]
3 years ago
15

When simplified completely, the product of a monomial and a monomial is sometimesalwaysnever a monomial

Mathematics
1 answer:
Vaselesa [24]3 years ago
7 0

Answer:

Always

Step-by-step explanation:

A mononomial's variable can only have exponents 0,1,2,3 etc  so the product will also be a mononomial.

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Help me out with this question (geometry)
Usimov [2.4K]

Answer:

Step-by-step explanation:

∠UST=2 m∠2

10x+10=2(6x-1)

10x+10=12x-2

12x-10x=10+2

2x=12

x=6

m∠UST=10×6+10=70°

4 0
3 years ago
The second digit after the decimal point in a number is 4. What is the value of 4 in the number? 1 thousandth 4 thousandths 1 hu
MrMuchimi

Answer:

D. 4 hundredths

Step-by-step explanation:

Hello!

The second digit after the decimal point is 4 is shown by

0.04

The points after a decimal point go tenths, hundredths, thousandths etc.

The answer is D. 4 hundredths

Hope this helps!

3 0
3 years ago
Read 2 more answers
The quadratic function f (x) = - 45x2 + 350x + 1,590 models the population of a city where x represents the number of years sinc
kramer

Answer: 2,215,000

Step-by-step explanation:

Given: The quadratic function f (x) = - 45x^2 + 350x + 1,590 models the population of a city where x represents the number of years since 2005.

To Find:  population of the city in 2010

We need to put x= 2010-2005 = 5

f (5) = - 45(5)^2 + 350(5) + 1,590\\\\=-1125+1750+1590 =2215

Hence, the estimated population of the city in 201 = 2215 thousands or 2,215,000 .

3 0
3 years ago
Kate bought $23.40 worth of two types of bird seed. Thistle bird seed sells for $1.60 per pound and wild bird seed sells for $0.
earnstyle [38]

Answer:

Ax+By=C and 7.5 lb of Thistle

Step-by-step explanation:

6 0
3 years ago
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
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