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omeli [17]
2 years ago
11

Dean is comparing prices on ground beef. Store A is selling 5 pounds of ground beef for $23.49. Store B is selling 8 pounds of g

round beef for $36.96.
Which store is offering the better deal on ground beef? Show your work.
Mathematics
2 answers:
Nastasia [14]2 years ago
6 0
The answer is answer A is the correct answer
pentagon [3]2 years ago
3 0

Answer: Store B

Step-by-step explanation:

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Serggg [28]
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-18n = 8m + 4
 /-18    /-18  /-18
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I'm not sure so yeah
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I need to solve this equation: x= r-h/ y<br> for h and then for r
ololo11 [35]
X= r-h/y
h= xy-r/-1
r= xy+h
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Lucy ate eight more than three times as many pretzels as Matt Lucy ate 56 pretzels how many pretzels did Matt eat
mr_godi [17]

Answer:

53

Step-by-step explanation:

56-3

5 0
2 years ago
which point on the graph shows the price of 3 pounds of meat ? use the formula y =3x to find y when x equals 3 ​
Anon25 [30]

Answer: y=9 (point B on the graph)

Step-by-step explanation:

y=3x

x=3

y=3(3) (3×3)

3x3=9

So 9(y)=3(3x)

Point B which is located at the 9 mark

5 0
2 years ago
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A flat circular plate has the shape of the region x squared plus y squared less than or equals 1x2+y2≤1. the​ plate, including t
vredina [299]

You're looking for the extreme values of x^2+3y^2+13x subject to the constraint x^2+y^2\le1.

The target function has partial derivatives (set equal to 0)

\dfrac{\partial(x^2+3y^2+13x)}{\partial x}=2x+13=0\implies x=-\dfrac{13}2

\dfrac{\partial(x^2+3y^2+13x)}{\partial y}=6y=0\implies y=0

so there is only one critical point at \left(-\dfrac{13}2,0\right). But this point does not fall in the region x^2+y^2\le1. There are no extreme values in the region of interest, so we check the boundary.

Parameterize the boundary of x^2+y^2\le1 by

x=\cos u

y=\sin u

with 0\le u. Then t(x,y) can be considered a function of u alone:

t(x,y)=t(\cos u,\sin u)=T(u)

T(u)=\cos^2u+3\sin^2u+13\cos u

T(u)=3+13\cos u-2\cos^2u

T(u) has critical points where T'(u)=0:

T'(u)=-13\sin u+4\sin u\cos u=\sin u(4\cos u-13)=0

(1)\quad\sin u=0\implies u=0,u=\pi

(2)\quad4\cos u-13=0\implies\cos u=\dfrac{13}4

but |\cos u|\le1 for all u, so this case yields nothing important.

At these critical points, we have temperatures of

T(0)=14

T(\pi)=-12

so the plate is hottest at (1, 0) with a temperature of 14 (degrees?) and coldest at (-1, 0) with a temp of -12.

4 0
2 years ago
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