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3 years ago

14

UCULI

2 answers:

Elis [28]3 years ago

6 0

Answer is B

UCULI
2 Points
When is a rhombus a square?
O
A. When its sides are parallel
B. When its angles are right angles
) C. When its angles are convex angles
O
D. When its sides are congruent
SUB

KatRina [158]3 years ago

5 0

Answer:

B. When its angles are right angles.

Step-by-step explanation:

A rhombus is a quadrilateral with all 4 sides congruent. The sides can be slanted, which would keep it from being a square. If you make all the sides of a rhombus straight vertically and horizontally so you have 4 right angles, you have a square. A square is a quadrilateral with all 4 sides AND all 4 angles congruent.

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Nataly_w [17]

Answer:

The equation that models the situation is $s_{f} = 4+\frac{v_{f}^{2}-2500}{64.348}$ .

Step-by-step explanation:

Let suppose that effects from air friction and Earth's rotation can be neglected, so that the softball can be modelled experiment a free fall, that is, an uniform accelerated motion due to gravity. From we know the initial velocity and position of the position and we can determine the final position of the ball as a function of the final velocity:

$v_{f}^{2} = v_{o}^{2}+2\cdot a \cdot (s_{f}-s_{o})$ (1)

Where:

$s_{o}$ , $s_{f}$ - Initial and final position of the softball, measured in feet.

$a$ - Acceleration, measured in meters per square second.

$v_{o}$ , $v_{f}$ - Initial and final velocities of the softball, measured in feet per second.

If we know that $v_{o} = 50\,\frac{ft}{s}$ , $a = -32.174\,\frac{m}{s^{2}}$ and $s_{o} = 4\,ft$ , then the equation that models the situation is:

$v_{f}^{2} = 2500-64.348\cdot (s_{f}-4)$

Then, we clear the final position of the softball:

$-\frac{v_{f}^{2}-2500}{-64.348} = s_{f}-4$

$s_{f} = 4+\frac{v_{f}^{2}-2500}{64.348}$

The equation that models the situation is $s_{f} = 4+\frac{v_{f}^{2}-2500}{64.348}$ .

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2 years ago

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

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A = 1/2 bh
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answer
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