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matrenka [14]
3 years ago
10

3 Gordon is selling popcorn that costs

Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
7 0

Answer:

30 more bags

Step-by-step explanation:

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General Mills, the manufacturer of Lucky Charms, claims that their new boxes hold 25% more cereal. If their original box held 50
shutvik [7]

Answer:

625

Step-by-step explanation:

basically, half of 500 is 250 halve that you get a quater of 500 (25%)

500+125=625

6 0
2 years ago
Given x = wavelength and y = frequency are inversely related, solve for the constant for the given type of
Natasha_Volkova [10]

Answer:

given

x = k 1/ y

so

k = x y

= 300,000 × 40

= 12,000,000

4 0
3 years ago
Read 2 more answers
Please help its due soon ill give brainlest
WITCHER [35]

Answer:

Rectangle, line segment, or point.

Step-by-step explanation:

If the plane intersects parallel to side, it can for a rectangle.

If the plane was slanted and touched the edge, then it will make a line segment.

If the plane was slanted and touched a corner, it will make a point.

6 0
3 years ago
Can someone help me with question 6 please
SIZIF [17.4K]
A) 4a+3w
b) 2b+h
c) 3(w+h)
d) (4a+3w)(2b+h) - 3(w+h)
just expand brackets for d and simplify
6 0
2 years ago
Read 2 more answers
Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |
dsp73

Looks like f(x,y)=x^2+y^2-x^2y+7.

f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1

f_y=2y-x^2=0\implies2y=x^2

  • If x=0, then y=0 - critical point at (0, 0).
  • If y=1, then x=\pm\sqrt2 - two critical points at (-\sqrt2,1) and (\sqrt2,1)

The latter two critical points occur outside of D since |\pm\sqrt2|>1 so we ignore those points.

The Hessian matrix for this function is

H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}

The value of its determinant at (0, 0) is \det H(0,0)=4>0, which means a minimum occurs at the point, and we have f(0,0)=7.

Now consider each boundary:

  • If x=1, then

f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4

which has 3 extreme values over the interval -1\le y\le1 of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

  • If x=-1, then

f(-1,y)=8-y+y^2

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

  • If y=1, then

f(x,1)=8

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

  • If y=-1, then

f(x,-1)=2x^2+8

which has 3 extreme values, but the previous cases already include them.

Hence f(x,y) has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

3 0
3 years ago
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