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Travka [436]
3 years ago
10

for quadrilateral ABCD, determine the most precise name for it. A(2,3), B(12,3), C(8,6) and D(5,6). Show your work and explain.

Mathematics
2 answers:
Vadim26 [7]3 years ago
7 0
1. Draw the points in the coordinate axes, as in the attached picture.

2. AB and CD are parallel, (both are parallel to the x-axis)

3. A is closer to the y axis than D and C is closer to the C closer than B

4. So combining 2 and 3, we conclude that ABCD is a trapezoid.

5. The remaining thing to check is whether the trapezoid is isosceles or not. For this we drop the heights to AB from points C and D, 

and see that the distances from the feets of these heights to the points A and B are not equal.


Answer: Trapezoid

mina [271]3 years ago
5 0
ABCD is a trapezium. Since line AB and CD do not vary in y, they are parallel. Lines AD and BC are not parallel, as AD increases in x, and BC decreases in x. A quadrilateral with only one set of parallel lines is a trapezium.
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7a-11=-2(3a-1) what is the work and answer
Anni [7]
The work is on the picture attached, but a = 1
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3 years ago
The graph 4x^2-4x-1 is shown. Use the grpah to find the estimates for the solutions of 4x^2-4x-1=0 and 4x^2 - 4x-1=2
Darina [25.2K]

Answer:

a) The estimates for the solutions of 4\cdot x^{2}-4\cdot x -1 = 0 are x_{1}\approx -0.25 and x_{2} \approx 1.25.

b) The estimates for the solutions of 4\cdot x^{2}-4\cdot x -1 = 2 are x_{1}\approx -0.5 and x_{2} \approx 1.5

Step-by-step explanation:

From image we get a graphical representation of the second-order polynomial y = 4\cdot x^{2}-4\cdot x -1, where x is related to the horizontal axis of the Cartesian plane, whereas y is related to the vertical axis of this plane. Now we proceed to estimate the solutions for each case:

a) 4\cdot x^{2}-4\cdot x -1 = 0

There are two approximate solutions according to the graph, which are marked by red circles in the image attached below:

x_{1}\approx -0.25, x_{2} \approx 1.25

b) 4\cdot x^{2}-4\cdot x -1 = 2

There are two approximate solutions according to the graph, which are marked by red circles in the image attached below:

x_{1}\approx -0.5, x_{2} \approx 1.5

5 0
3 years ago
In triangle $ABC$, let angle bisectors $BD$ and $CE$ intersect at $I$. The line through $I$ parallel to $BC$ intersects $AB$ and
Umnica [9.8K]

Answer:

41

Step-by-step explanation:

If you work through a series of obscure calculations involving area and the radius of the incircle, they boil down to a simple fact:

... For MN║BC, perimeter ΔAMN = perimeter ΔABC - BC = AB+AC

.. = 17+24 = 41

_____

Wow! Thank you for an interesting question with a not-so-obvious answer.

_____

<em>A little more detail</em>

The point I that you have defined is the incenter—the center of an inscribed circle in the triangle. Its radius is the distance from I to any side, such as BC, for example.

If we use "Δ" to represent the area of the triangle and "s" to represent the semi-perimeter, (AB+BC+AC)/2, then the incircle has radius Δ/s. The area Δ can be computed from Heron's formula by ...

... Δ = √(s(s-a)(s-b)(s-c)) . . . . where a, b, c are the side lengths

For this triangle, the area is Δ = √38480 ≈ 196.1632 units². That turns out to be irrelevant.

The altitude to BC will be 2Δ/(BC), so the altitude of ΔAMN = (2Δ/(BC) -Δ/s). Dividing this by the altitude to BC gives the ratio of the perimeter of ΔAMN to the perimeter of ΔABC, which is 2s.

Putting these ratios and perimeters together, we get ...

... perimeter ΔAMN = (2Δ/(BC) -Δ/s)/(2Δ/(BC)) × 2s

... = (2/(BC) -1/s) × BC × s = 2s -BC

... perimeter ΔAMN = AB +AC

8 0
3 years ago
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Oksana_A [137]
V= s^3  
a=486/6=81
s=sqrt(81)=9
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the volume of the cube is 729cm^3
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