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3 years ago

10

Which equation shows the relationship between the x and y values? Its easy I learned this in 1st grade, but I'm to lazy to even

try. *Please Help* :D

1 answer:

Nata [24]3 years ago

8 0

Y= 3x-4 Is the rule for table to check .

Take first one

7,17

17= 3(7)-4

17=21-4

17=17 it checks

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PLEASE SOMEONE TO HELP ME WHAT ARE THE CORRECT ANSWERS AND IM GIVING BRAINLY

GrogVix [38]

Answer:

10 times greater

Step-by-step explanation:

When you write these as numbers, you get:

800,000,000 (8 hundred million place)

80,000,000 (8 ten million place)

Then divide the greater number by the less number:

800,000,000/800,000,00

make this easier by eliminating 0's:

80<u>0,000,000</u>/8<u>0,000,000</u>

80/8

10

10 times greater is your answer

Hope this helps!

4 0

3 years ago

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Simplify the expression write the answer in scientific notation (9×10^7) <br>(7×10^9)

aliina [53]

$\text{The scientific notation}\ a\cdot10^k,\ \text{where}\ 1\leq a$

6 0

3 years ago

On her way to work each morning sophia purchase a small cup of coffee for 4.25 from the coffee shop

steposvetlana [31]

What do you want us to answer?

5 0

3 years ago

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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

Please help me solve this correct answer gets brainlist

julia-pushkina [17]

Answer:

hyp²= opp² + adj²

hyp²= 12² +4²

hyp²= 144+16

hyp²=160

hyp=√160

hyp= 4√10

8 0

3 years ago

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