So you must find individual prices.
Start by doing the price($2.00)divided by the amount in box(8)
you should get 0.25 or 25 cents per chocolate
Answer:
y = 9x/5 + 50
Step-by-step explanation:
We are represent the information as coordinate (x,y)
If the cost for an order of 100 kilograms of steel bars is $230, this is expressed as (100, 230)
Also if the cost for an order of 150 kilograms of steel bars is $320, this is expressed as;
(150, 320)
Find the equation of a line passing through the points. The standard form of the equation is expressed as y = mx+c
m is the slope
c is the intercept
Get the slope;
m = y2-y1/x2-x1
m = 320-230/150-100
m = 90/50
m = 9/5
Get the y-intercept by substituting m = 9/5 and any point say (100, 230) into the expression y = mx+c
230 = 9/5(100)+c
230 = 9(20)+c
230 = 180 + c
c = 230-180
c = 50
Get the required equation
y = mx+c
y = 9/5 x + 50
Hence an equation for the cost of an order of steel bars (y) in terms of the weight of steel bars ordered (x) is y = 9x/5 + 50
Answer:

Step-by-step explanation:
we would like to solve the following equation for x:

to do so isolate
to right hand side and change its sign which yields:

simplify Substraction:

get rid of only x:

simplify addition of the left hand side:

divide both sides by q+p Which yields:

cross multiplication:

distribute:

isolate -pq to the left hand side and change its sign:

rearrange it to standard form:

now notice we end up with a <u>quadratic</u><u> equation</u> therefore to solve so we can consider <u>factoring</u><u> </u><u>method</u><u> </u><u> </u>to use so
factor out x:

factor out q:

group:

by <em>Zero</em><em> product</em><em> </em><em>property</em> we obtain:

cancel out p from the first equation and q from the second equation which yields:

and we are done!
Answer:
The answer is B.
Step-by-step explanation:
Step-by-step explanation:
Discriminant : k² - 4(12)(27) < 0
So k² - 1296 < 0, -36 < k < 36.