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Sever21 [200]
4 years ago
9

mark used a graph to compare the wages from two restaurans jobs.a dishwasher earns $7.25 per hour,while a waiter earns $3.75 per

hour addition to $25 per day
Mathematics
1 answer:
Alla [95]4 years ago
4 0
Y=14.50

Y=12     

hope this helps.



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A contractor hires 4 workers for a construction project. Each worker is paid $12 per hour. The contractor also buys 20 cubic yar
vovangra [49]

Answer:

The answer is C.

Step-by-step explanation:

4 x 12 is how much money per worker per hour. 20 x 160 is how much the cost for the concrete. Adding them gets you the total cost.

Hope this helped!

7 0
3 years ago
Read 2 more answers
Plz help ill give u brainlist
belka [17]
C. 5x+7=22 since diameter is equal to double the radius. radius=11 so diameter equals 22. then just make 5x+7 = to 22
6 0
4 years ago
Read 2 more answers
I’m being timed !!!
Charra [1.4K]

Step-by-step explanation:

I dont know the answer but I know you should make a graph for it and it will tell u the answer I will try and take a photo

3 0
3 years ago
Use lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f(x,y = xyz; x^
Snezhnost [94]
I'm assuming the constraint involves some plus signs that aren't appearing for some reason, so that you're finding the extrema subject to x^2+2y^2+3z^2=96.

Set f(x,y,z)=xyz and g(x,y,z)=x^2+2y^2+3z^2-96, so that the Lagrangian is

L(x,y,z,\lambda)=xyz+\lambda(x^2+2y^2+3z^2-96)

Take the partial derivatives and set them equal to zero.

\begin{cases}L_x=yz+2\lambda x=0\\L_y=xz+4\lambda y=0\\L_z=xy+6\lambda z=0\\L_\lambda=x^2+2y^2+3z^2-96=0\end{cases}

One way to find the possible critical points is to multiply the first three equations by the variable that is missing in the first term and dividing by 2. This gives

\begin{cases}\dfrac{xyz}2+\lambda x^2=0\\\\\dfrac{xyz}2+2\lambda y^2=0\\\\\dfrac{xyz}2+3\lambda z^2=0\\\\x^2+2y^2+3y^2=96\end{cases}

So by adding the first three equations together, you end up with

\dfrac32xyz+\lambda(x^2+2y^2+3z^2)=0

and the fourth equation allows you to write

\dfrac32xyz+96\lambda=0\implies \dfrac{xyz}2=-32\lambda

Now, substituting this into the first three equations in the most recent system yields

\begin{cases}-32\lambda+\lambda x^2=0\\-32\lambda+2\lambda y^2=0\\-32\lambda+3\lambda z^2=0\end{cases}\implies\begin{cases}x=\pm4\sqrt2\\y=\pm4\\z=\pm4\sqrt{\dfrac23}\end{cases}

So we found a grand total of 8 possible critical points. Evaluating f(x,y,z)=xyz at each of these points, you find that f(x,y,z) attains a maximum value of \dfrac{128}{\sqrt3} whenever exactly none or two of the critical points' coordinates are negative (four cases of this), and a minimum value of -\dfrac{128}{\sqrt3} whenever exactly one or all of the critical points' coordinates are negative.
6 0
3 years ago
Olivia and her three siblings bring a sack lunch to school each day, consisting of a bagel, an apple, a cookie, and a juice box.
Vikki [24]
Wow OK so the first number I came to was 72 so that's
bagels 6x12=72
apples 8x9=72
cookies 12x6=72
juice boxes 9x8=72

72/4=18 so she could make their lunch for 18 days
7 0
3 years ago
Read 2 more answers
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