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xenn [34]
3 years ago
7

What is the perimeter of polygon lmnpq?

Mathematics
1 answer:
natita [175]3 years ago
8 0

Answer:

The total distance of the outer sides of a closed figure is known as the perimeter. It is the total length of all sides of a polygon. Perimeter = Sum of all sides.

Formula:

The sum of interior angles of a polygon with “n” sides =180°(n-2) Number of diagonals of a “n-sided” polygon = [n(n-3)]/2. The measure of interior angles of a regular n-sided polygon = [(n-2)180°]/n. The measure of exterior angles of a regular n-sided polygon = 360°/n.

Step-by-step explanation:

hope it helps...mark me as a brainlist

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Olivia borrows $50 from her mom to buy some art supplies. Olivia spends $32.15 on art supplies and gives the change back to her
frutty [35]

Mom's view is that she loses 50 dollars, but gains back 17.85 dollars as change from her daughter.

Olivia's view is that she gained 50 dollars from her mom, spent 32.15 dollars, then gave back the 17.85 change to her mom, so she didn't gain any money.


Mom's point of view: -$50 + $17.85 and Mom's point of view: -$50 + $17.85 are the correct answers, but technically, Olivia still owes her mom 32.15 dollars.

6 0
3 years ago
Hey, please help me, I'm fairly confused with this question.
Sati [7]

Answer:

10 times.

Step-by-step explanation:

The number in the tenths place is 90. The number in the ones place is 9. 9 times 10 is 90.

5 0
3 years ago
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77julia77 [94]
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3 years ago
3∙(a+x), if a=8; x=−10
Oksi-84 [34.3K]
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6 0
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Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
melamori03 [73]

Answer:

6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs AB=3, BC=4, and AC=5. The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base b and height h is given by A=\frac{1}{2}bh. Therefore, the area of this right triangle is:

A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}

The other triangle is a bit trickier. Triangle \triangle ADC is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

A=\sqrt{s(s-a)(s-b)(s-c)}, where a, b, and c are three sides of the triangle and s is the semi-perimeter (s=\frac{a+b+c}{2}).

The semi-perimeter, s, is:

s=\frac{5+5+4}{2}=\frac{14}{2}=7

Therefore, the area of the isosceles triangle is:

A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}

Thus, the area of the quadrilateral is:

6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}

4 0
3 years ago
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