What is the area of the trapezoid below

[WILL PICK FIRST CORRECT ANSWER AS BRAINLIEST]

riadik2000 [5.3K]

F(3) = 6 - 3 = 3

answer 3

4 0

3 years ago

Read 2 more answers

8/9 + 4/7 = Can you get the Anwar for this sum

attashe74 [19]

here. I don't know if its right but i hope it is.

7 0

3 years ago

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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.

_____

A little more detail

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

An experiment consists of rolling two fair number cubes. What is the probability that the sum of the two numbers will be >3

liraira [26]

Answer: $\bold{\dfrac{5}{12}}$

Step-by-step explanation:

greater than 3 means the sum is either 4, 5, or 6.

There are 15 permutations that meet the criteria:

(1, 3), (1, 4), (1, 5), (1,6)

(2, 2), (2, 3), (2, 4), (2, 5), (2,6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3,6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4,6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5,6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6,6)

There are 6²= 36 total permutations for 2 dice.

$Probability=\dfrac{\#\ of\ permutations> 3}{total\ \#\ of\ permutations}\\\\.\qquad \qquad =\dfrac{15}{36}\\\\.\qquad \qquad =\dfrac{5}{12}\\\\$

3 0

3 years ago

Its on the picture, plz go fast

Naily [24]

Answer:

Elimination

Step-by-step explanation:

You would not do substitution since it would not work so you do elimination since the 4y and -4y eliminate then you just solve for it.

7 0

2 years ago