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expeople1 [14]
3 years ago
5

Which pairs of polygons are similar? Select each correct answer. Two trapezoids. The trapezoid on the left has horizontal bases

at the top and bottom. The left leg is perpendicular to the bases. The right leg is slanted. The top right angle has a single arc and the bottom right angle has a pair of arcs. The top side is 7.2 inches, the right side is 27 inches, the bottom side is 18 inches, and the left side is 10.8 inches. The trapezoid on the right also has horizontal bases. The right leg is parallel to the bases. The left leg is slanted. The top left angle has a single arc. The bottom left angle has a pair of arcs. The top side is 4 inches, the right side is 6 inches, the bottom side is 10 inches, and the left side is 15 inches. Two rectangles. The larger rectangle has side lengths of 14 millimeters and 18 millimeters. The smaller rectangle has side lengths of 7 millimeters and 10 millimeters. Two triangles. In the smaller triangle, all three sides are 5 feet long. In the larger triangle, all three sides are 8 feet long. Two right triangles. One of the acute angles in each triangle has a single congruent arc. In the larger triangle, the sides measure 24 centimeters, 32 centimeters, and 40 centimeters. In the smaller triangle, the sides measure 18 centimeters, 24 centimeters, and 30 centimeters.
Mathematics
2 answers:
balu736 [363]3 years ago
8 0

Answer:

*thinking*

Step-by-step explanation:

Yea Ima need you to add a picture of the pairs then I should be able to help. Ill follow so ill see when you update

Murljashka [212]3 years ago
4 0

Answer:

i need picture to help

Step-by-step explanation:

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X\mleft(t\mright)=\frac{2}{3}t^3-\frac{9}{2}t^2-18t

The velocity function is the derivative of the position:

\begin{gathered} V(t)=\frac{2}{3}(3t^2)-\frac{9}{2}(2t)-18 \\ \text{Simplifying:} \\ V(t)=2t^2-9t-18 \end{gathered}

The particle will be at rest when the velocity is 0, thus we solve the equation:

2t^2-9t-18=0

The coefficients of this equation are: a = 2, b = -9, c = -18

Solve by using the formula:

t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}

Substituting:

\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}

We have two possible answers:

\begin{gathered} t=\frac{9+15}{4}=6 \\ t=\frac{9-15}{4}=-\frac{3}{2} \end{gathered}

We only accept the positive answer because the time cannot be negative.

Now calculate the position for t = 6:

undefined

6 0
1 year ago
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Step-by-step explanation:

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Step-by-step explanation:

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