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kaheart [24]
3 years ago
11

(NEED HELP ASAP!!!) Jamie unfolded a cardboard box. The figure of the unfolded box is shown below: A long rectangle divided alon

g the length into 4 equal squares is shown. Along the upper edge length of the second square is shown another square, and along the lower edge length of the fourth square is shown another square. The first square from the left has two adjacent sides labeled as 4 inches each. Which expression can be used to calculate the area of cardboard, in square inches, that was used to make the box? 6 x 4 x 4 4 x 6 x 6 6 x 2 x 2 2 x 6 x 6
Mathematics
1 answer:
Oksanka [162]3 years ago
4 0

Answer:

6 x 4 x 4 Hope I get u right<3

Step-by-step explanation:

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15. JAGUARUNDI A jaguarundi springs from a fence post to swat at a low flying bird.
sukhopar [10]

Answer:

The discriminant, -14.71 < 0, shows that there is no solution of the equation, h = -16·t² + 22.3·t + 2, at the line 'h = 10' feet, therefore, the Jaguarundi height as she springs will not be up to 10 feet and therefore she will not reach the bird

Step-by-step explanation:

From the question, the equation that models the Jaguarundi's height, 'h', ma be written approximately as follows;

h = -16·t² + 22.3·t + 2

Where;

t = The time (duration) in seconds

The discriminant of the equation a·x² + b·x + c = 0 is  b² - 4·a·c

When h = 10, we have;

10 = -16·t² + 22.3·t + 2

∴ 0 = -16·t² + 22.3·t + 2 - 10 = -16·t² + 22.3·t - 8

The discriminant of the given quadratic equation is given as follows;

The discriminant = 22.3² - 4 × (-16 × (-8)) = -14.71 < 0

Therefore, the function, h = -16·t² + 22.3·t + 2 has no real root at h = 10

The parabola does not reach or pass through the line h = 10 which is the height at which the bird is flying.

The Jaguarundi will not reach the bird flying at the height of 10 feet.

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3 years ago
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I’m not quite sure how to solve this any help would work!
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Answer:

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Step-by-step explanation:

11. Circumcentre

Your points form the vertices of a right triangle (see image below).

The circumcentre (O) of a right triangle must be at the midpoint of the hypotenuse.

Your hypotenuse has the ends at (-3,4) and (1,1), so

O = ((-3 + 1)/2),(4 + 1)/2))

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The orthocentre (H) of a right triangle is the vertex at the right angle.

Your right angle is at (1, 4), so

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