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Vika [28.1K]
4 years ago
11

The figure below shows a quadrilateral ABCD with diagonal BD bisecting angle ADC.

Mathematics
2 answers:
Scorpion4ik [409]4 years ago
5 0
So based on the given figure above which is quadrilateral ABCD, the equation that is true would be the third option: angle ABD = angle CBD. Based on the SAS Postulate, i<span>f two sides and the included angle of a triangle are congruent to the corresponding sides and angle of another triangle , then the triangles are congruent. Hope that this answer helps.</span>
Ad libitum [116K]4 years ago
3 0

It is the third one: Angle ABD = Angle CBD.

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Will give brainliest answer
timofeeve [1]

Answer:

No

Step-by-step explanation:

This is because when a triangle has side lengths a,b, and c, a+b>c.

But 6.0+4.2 is not greater than 12.2, so this not a triangle.

5 0
3 years ago
Someone please explain the solution to this :
juin [17]

Answer: 17.5 units.

Step-by-step explanation:

The triangle LAY is a similar triangle to LYW. It has the same angle measures, but different side lengths. The hypotenuse of LAY is the line LY, while the hypotenuse of LYW is the line LW.

We can use the line LY to find LW.

Use the Pythagorean Theorem to find LY:

(3.5)^2 + 7^2 = 61.25

So, LY has a length of \sqrt{61.25}.

You can then use proportions to find the distance of LW.

The line LA is the base of the smaller triangle, and the line LY is the base of the larger triangle.

The line LY is the hypotenuse of the smaller triangle, and the line LW is the hypotenuse of the larger triangle.

\frac{3.5}{\sqrt{61.25} } =\frac{\sqrt{61.25} }{y}

You can then cross multiply to get this equation:

61.25 = 3.5y

So, the line LW has a length of 17.5.

3 0
3 years ago
Explain to a classmate why you can use any two sides of a right triangle to find the third side
STatiana [176]
Because, all sides are equal
5 0
4 years ago
Read 2 more answers
What is the simplified expression in standard form
Schach [20]

Answer:

\boxed{\sf - 2 {p}^{2}  - 11p - 35}

Step-by-step explanation:

\sf \implies - 2 {(p + 4)}^{2}  - 3 + 5p \\  \\  \sf \implies  - 2( {p}^{2}  + 2(p)(4) +  {4}^{2} ) - 3 + 5p \\  \\  \sf \implies  - 2( {p}^{2}  + 8p + 16) - 3 + 5p \\  \\  \sf \implies (( - 2) \times  {p}^{2} ) + (( - 2) \times 8p) + ( ( - 2) \times 16) - 3 + 5p \\  \\  \sf \implies  (- 2 {p}^{2} ) + ( - 16p ) + (- 32) - 3 + 5p \\  \\  \sf \implies  - 2 {p}^{2}  - 16p - 32 - 3 + 5p \\  \\  \sf \implies  - 2 {p}^{2} +   (- 16p + 5p) + ( - 32 - 3) \\  \\  \sf \implies  - 2 {p}^{2}  - 11p - 35

4 0
4 years ago
Explain why you would make one of the addends a tens number when solving an addition problem.
jeka57 [31]

Since we can add a number to a tens number easily, we make one of the addends a tens number when solving an addition problem.

Example 1:

Add 9 and 6.

9 + 6 = 9 + (1 + 5)

        = (9 + 1) + 5

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Note that adding 10 and 5 is much easier than adding 9 and 6.

Example 2:

Add 18 and 16.

18 + 16 = 18 + (2 + 14)

        = (18 + 2) + 14

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Note that adding 20 and 14 is much easier than adding 18 and 16.

5 0
4 years ago
Read 2 more answers
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