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puteri [66]
4 years ago
8

HELP!! Fast! Plz! Is triangle ABC similar to DEC? Explain how you know.

Mathematics
2 answers:
Alekssandra [29.7K]4 years ago
6 0
Angle B=46° Angle B=72° Angle A= 62°
Since they share Angle C they must be equal.
astraxan [27]4 years ago
4 0
Because where the lines make the x symbol the measurement on the other side has to be the same on the opposite side.
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Tenemos 2 rectangulos, el primero mide 15 centimetros de largo y 4 centimetros de ancho al formar el segundo rectangulo se dupli
sergey [27]

Answer:

El segundo rectangulo va medir 30 centimetros de largo y 8 centimetros de ancho.

Step-by-step explanation:

Duplicar las medidas significa que vamos multiplicar las medidas por 2.

15 cm × 2= 30 cm

4 cm × 2= 8 cm

Espero que te ayude!

8 0
3 years ago
An example of a rectangular Prism can be: A globe or ball O Fish tank o Bowl Can​
garik1379 [7]

Answer:

Fish Tank

Step-by-step explanation:

This of a rectangular prism as a cardboard box. It has rectangles on the top and the bottom. The fish tank is somewhat similar to this so it will be a rectangular prism.

3 0
3 years ago
Find the area of object and what is method
Alborosie

It would be...

(18 +28) divided by two

then take that and multiply it by the height - 5

the 6 in this equation doesn't matter because it is not the height of the shape

8 0
4 years ago
Plss I need this answer....pls help me. Explain this question and answer it.​
AVprozaik [17]

Answer:

Figure three is the odd one out

Step-by-step explanation:

it is different from the other figures. The other figures are all the same, they're just rotated. Figure three is different from all of them though

3 0
3 years ago
At which points are the tangents drawn to the ellipse x 2 + y 2 = [ a ] x + [ a ] y parallel to
In-s [12.5K]

The given equation of the ellipse is x^2 + y^2 = 2 x + 2 y

At tangent line, the point is horizontal with the x-axis therefore slope = dy / dx = 0

<span>So we have to take the 1st derivative of the equation then equate dy / dx to zero.</span>

x^2 + y^2 = 2 x + 2 y

x^2 – 2 x = 2 y – y^2

(2x – 2) dx = (2 – 2y) dy

(2x – 2) / (2 – 2y) = 0

2x – 2 = 0

x = 1

 

To find for y, we go back to the original equation then substitute the value of x.

x^2 + y^2 = 2 x + 2 y

1^2 + y^2 = 2 * 1 + 2 y

y^2 – 2y + 1 – 2 = 0

y^2 – 2y – 1 = 0

Finding the roots using the quadratic formula:

y = [-(- 2) ± sqrt ( (-2)^2 – 4*1*-1)] / 2*1

y = 1 ± 2.828

y = -1.828 , 3.828

 

<span>Therefore the tangents are parallel to the x-axis at points (1, -1.828) and (1, 3.828).</span>

3 0
4 years ago
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