The function y varies inversely as x, and y = 7 when x = 5. What is the value of y when x = 14?

Sandy has box of chocolates. Box cost $2.00. In box has 8 chocolates. How much does each chocolate cost.

Lorico [155]

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

5 0

2 years ago

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The cost for an order of 100 kilograms of steel bars is $230. The cost for an order of 150 kilograms of steel bars is $320. Writ

luda_lava [24]

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

4 0

2 years ago

Solve <img src="https://tex.z-dn.net/?f=%5Csf%20%5Cdfrac%7B1%7D%7Bp%7D%20%2B%20%5Cdfrac%7B1%7D%7Bq%7D%20%2B%20%5Cdfrac%7B1%7D

Nostrana [21]

Answer:

$\displaystyle \begin{cases} \displaystyle {x} _{1} = - p \\ \displaystyle x _{2} = - q \end{cases}$

Step-by-step explanation:

we would like to solve the following equation for x:

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{p + q + x}$

to do so isolate $\frac{1}{x}$ to right hand side and change its sign which yields:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{1}{p + q + x} - \frac{1}{x}$

simplify Substraction:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{x - (q + p + x)}{x(p + q + x)}$

get rid of only x:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{ - (q + p )}{x(p + q + x)}$

simplify addition of the left hand side:

$\displaystyle \frac{q + p}{pq} = \frac{ - (q + p )}{x(p + q + x)}$

divide both sides by q+p Which yields:

$\displaystyle \frac{1}{pq} = \frac{ -1}{x(p + q + x)}$

cross multiplication:

$\displaystyle x(p + q + x) = - pq$

distribute:

$\displaystyle xp + xq + {x}^{2} = - pq$

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

$\displaystyle xp + xq + {x}^{2} + pq = 0$

rearrange it to standard form:

$\displaystyle {x}^{2} + xp + xq + pq = 0$

now notice we end up with a quadratic equation therefore to solve so we can consider factoring method to use so

factor out x:

$\displaystyle x( {x}^{} + p ) + xq + pq = 0$

factor out q:

$\displaystyle x( {x}^{} + p ) +q (x + p)= 0$

group:

$\displaystyle ( {x}^{} + p ) (x + q)= 0$

by Zero product property we obtain:

$\displaystyle \begin{cases} \displaystyle {x}^{} + p = 0 \\ \displaystyle x + q= 0 \end{cases}$

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

$\displaystyle \begin{cases} \displaystyle {x}^{} = - p \\ \displaystyle x = - q \end{cases}$

and we are done!

3 0

2 years ago

If there is an increase in the money supply that causes prices to rise and

andrew-mc [135]

Answer:

The answer is B.

Step-by-step explanation:

5 0

3 years ago

For what real values of does the quadratic 12x^2+k*x+27=0 have nonreal roots

icang [17]

Step-by-step explanation:

Discriminant : k² - 4(12)(27) < 0

So k² - 1296 < 0, -36 < k < 36.

3 0

2 years ago

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