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valkas [14]
3 years ago
10

If each muffin contains the same amount of cornmeal, how many kilograms of cornbread are in each corn muffin

Mathematics
2 answers:
Diano4ka-milaya [45]3 years ago
5 0
21 cornbread are in each corn muffin
Nookie1986 [14]3 years ago
5 0
Twenty One pieces of cornbread are in each corn muffin.
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For the function g(x)=3x^2-3x-9 what is the value of g(4)
Annette [7]
G(4) = 3x^2 - 3x - 9

Since the g(4) would be g(x) if you didn't know the number. So, substitute all of the x's for 4.

g(4) = 3(4)^2 - 3(4) - 9

g(4) = 3(16) - 12 - 9

g(4) = 48 - 3

g(4) = 45

I hope this helps you!
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3 years ago
A sugar solution currently measures 250 degrees Fahrenheit. The temperature of the solution increases 5 degrees per minute. The
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Let, the number of minutes = x
It  would be: 310 < 250 + 5x < 338

In short, Your Answer would be Option A

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5 0
3 years ago
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An object with a mass of 4 kg is hanging from an axle with a radius of 36 cm. if the wheel attached to the axle has a radius of
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7 0
3 years ago
The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?
Nuetrik [128]

Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

A(-6, 2)) = (x_1, y_1)

B(2, 6) = (x_2, y_2)

AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}

AB = \sqrt{(8)^2 + (4)^2}

AB = \sqrt{64 + 16} = \sqrt{80}

AB = 8.9 (nearest tenth)

Distance between B(2, 6) and C(7, 1):

BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

B(2, 6) = (x_1, y_1)

C(7, 1) = (x_2, y_2)

BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}

BC = \sqrt{(5)^2 + (-5)^2}

BC = \sqrt{25 + 25} = \sqrt{50}

BC = 7.1 (nearest tenth)

Distance between C(7, 1) and D(3, -5):

CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

C(7, 1) = (x_1, y_1)

D(3, -5) = (x_2, y_2)

CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}

CD = \sqrt{(-4)^2 + (-6)^2}

CD = \sqrt{16 + 36} = \sqrt{52}

CD = 7.2 (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

D(3, -5) = (x_1, y_1)

A(-6, 2) = (x_2, y_2)

DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}

DA = \sqrt{(-9)^2 + (7)^2}

DA = \sqrt{81 + 49} = \sqrt{130}

DA = 11.4 (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

8 0
3 years ago
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