Let H = Hailey, M = Megan, B = Bailey

H + M + B = 458
167 + M + B = 458
M + B = 291

152 ≤ M ≤ 164
If M = 152, B = 139
If M = 164, B = 127

Thus,
$127 ≤ B ≤ 139

4 0

3 years ago

The table show the relationship of how many pounds of cherries are needed to make a certain number of pies: pies 6 12 18 cherrie

Nataly_w [17]

M=dc/dp=4/6=4/6=2/3

c(p)=2p/3 +b using (6,4)

4=2(6)/3+b

4=4+b, b=0 so

c(p)=2p/3

(the number of cherries needed as a function of the number of pies)

3 0

3 years ago

Can someone explain the transitive property?

Fynjy0 [20]

Answer:

if a=b and b=c, then a=c

Step-by-step explanation:

Basically the transitive property states that is a number, a, equals another number, b, a = b, and the other number, b, equals a different number, c, b = c, then technically since a and b are equal, and b and c are equal, a could also equal c, a = c.

6 0

3 years ago

A large square consists of four identical rectangles and a small square. The area of the larg square is 49 cm2 and the length of

trapecia [35]

Answer:

The area of the small square is 1 cm^2

Step-by-step explanation:

The large square consist in four identical rectangles and one small square.

Then the area of the small square will be equal to the difference between the area of the large square and the areas of the rectangles.

Because we have 4 equal rectangles, if R is the area of one rectangle, and S is the area of the large square, the area of the small square will be:

area = S - 4*R

We know that the area of the large square is 49 cm^2

Then:

S = 49cm^2

Remember that the area of a square of side length K is:

A = K^2

Then the side length of the large square is:

K^2 = 49 cm^2

K = √(49 cm^2) = 7cm

And we know that the diagonal of one rectangle is 5cm.

Remember that for a rectangle of length L and width W, the diagonal is:

D = √(L^2 + W^2)

Then:

D = √(L^2 + W^2) = 5cm

And for how we construct this figure, we must have that the length of the rectangle plus the width of the rectangle is equal to the side length of the large square, then:

L + W = 7cm

L = (7cm - W)

Replacing this in the diagonal equation, we get:

√((7cm - W)^2 + W^2) = 5cm

(7cm - W)^2 + W^2 = (5cm)^2 = 25cm^2

49cm^2 - 14cm*W + W^2 + W^2 = 25cm^2

2*W^2 - 14cm*W + 49cm^2 = 25cm^2

2*W^2 - 14cm*W + 49cm^2 - 25cm^2 = 0

2*W^2 - 14cm*W + 24cm^2 = 0

We can solve this for W using the Bhaskara's formula, the solutions are:

$W = \frac{-(-14cm) \pm \sqrt{(-14cm)^2 - 4*2*(24cm^2)} }{2*2} = \frac{14cm \pm 2cm}{4}$

Then we have two solutions, and we only need one (because the length will have the other value)

We can take:

W = (14 cm + 2cm)/4 = 4cm

Then using the equation:

L + W = 7cm

L + 4cm = 7cm

L = 7cm - 4cm = 3cm

L = 3cm

Now remember that the area of one rectangle of length L and width W is:

R = L*W

Then the area of one of these rectangles is:

R = 4cm*3cm = 12cm^2

Now we can compute the area of the small square:

area = S - 4*R = 49cm^2 - 4*12cm^2 = 1cm^2

The area of the small square is 1 cm^2

3 0

3 years ago