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2 years ago

Can someone help me on letters B and C?

Mathematics

1 answer:

skelet666 [1.2K]2 years ago

4 0

Step-by-step explanation:

if he is just talking about the length than it is,

4(13x + 8)

of he is talking about overall length he swam than,

4(13x + 8) + 3(7x - 2)

if x = 12

than

perimeter = 2l + 2b

so, after inserting the values we got,

→ 2[13(12) + 8] + 2[7(12) - 2]

→ 2 (156 + 8) + 2(84 - 2)

→ 2 × 164 + 2 × 82

→ 328 + 164

→ 492 ft. ← perimeter

and yeah, according to me

a is wrong as,

formula of perimeter = 2l + 2b or 2(l+b)

so, the answer should be,

2(13x + 8) + 2(7x - 2) or 2 [ (13x + 8) + (7x - 2)

hope this answer helps you dear...take care and may u have a great day ahead!

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Find the area a of the triangle whose sides have the given lengths. a = 20, b = 15, c = 25 a =?

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The area of a triangle with sides a = 20, b = 15, and c = 25 is 150.

The sides of the triangle are given as a = 20, b = 15, and c = 25.

We will use Hero's formula to find the area of this triangle.

<h3>What is Heron's formula?</h3>

It is a three-face polygon that consists of three edges and three vertices.

We use Heron's formula to find the area of a triangle with 3 sides:

Herons formula:

Area of a triangle = $\sqrt{s(s-a)(s-b)(s-c)}\\$

Where a, b, and c are sides of a triangle.

And s = semi perimeter of a triangle.

s = $\frac{a+b+c}{2}$

If the sum of two sides of a triangle is greater than the third side of a triangle then the sides of a triangle are true.

Let the given sides be:

a = 20, b = 15 and c = 25.

(20 + 15) > 25

(20 + 25) > 15

(15 + 25) > 20 so the given sides are true.

Now,

Semi perimeter of the triangle:

s = (a+b+c) / 2

s = (20+15+25) / 2

s = 60 / 2

s = 30

Putting s = 30 in the area of the triangle.

we get,

Area of the triangle = $\sqrt{s(s-a)(s-b)(s-c)}\\$

Area of the triangle = $\sqrt{30(30-20)(30-15)(30-25)}\\\\\sqrt{30\times10\times15\times5}\\\\\sqrt{22500}\\\\150$

Thus, the area of a triangle is 150.

Learn more about the Area of triangles here:

brainly.com/question/11952845

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What set of reflections would carry trapezoid ABCD onto itself? (1 point) Trapezoid ABCD is shown. A is at negative 5, 1. B is a

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y=x, x axis, y=x, y axis

Step-by-step explanation:

If a point A(x, y) is reflected along the y = x line, the new coordinates would be A'(y, x)

If a point A(x, y) is reflected across the x axis, the new coordinates would be A'(x, -y)

If a point A(x, y) is reflected along the y axis, the new coordinates would be A'(-x, y)

Firstly reflect Trapezoid ABCD along the y = x line to give:

A (-5, 1) ⇒ A'(1, -5). B (-4, 3) ⇒ B'(3, -4). C (-2, 3) ⇒C'(3, -2). D (-1, 1) ⇒ D'(1, -1)

Secondly reflect along the x axis to give:

A'(1, -5)⇒ A"(1,5). B'(3, -4)⇒ B"(3,4), C'(3, -2)⇒C"(3, 2), D'(1, -1) ⇒ D"(1, 1)

Thirdly reflect along the y = x line to give:

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Lastly reflect along the y axis to give:

A"'(5, 1) ⇒ A""(-5, 1), B"'(4, 3) ⇒ B""(-4, 3), C"'(2, 3) ⇒ C""(-2, 3), D"'(1, 1) ⇒ D""(-1.1)

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