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4 years ago

1. Find the area and perimeter of the triangle. 16 cm 20.9 am 27 cm

Mathematics

1 answer:

Andrei [34K]4 years ago

7 0

Answer:

Perimeter = 63.9 cm.

Area = 166.96 cm^2 to the nearest hundredth.

Step-by-step explanation:

The perimeter = 16+20.9+27 = 63.9 cm.

To find the area you can use Heron's Formula.

Area = √ [ s(s-a)(s-b)(s-c)] where s = the semi-perimeter and a, b and c are the lengths of the 3 sides.

s = 63.9 / 2 = 31.95.

Area of the triangle = √ (31.95(31.95-16)(31.95-20.9)(31.95-27)]

= 166.96 cm^2.

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3 years ago

16) Please help with question. WILL MARK BRAINLIEST + 10 POINTS.

Katyanochek1 [597]

We will use the sine and cosine of the sum of two angles, the sine and consine of $\frac{\pi}{2}$ , and the relation of the tangent with the sine and cosine:

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\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta$

$\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0$

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If you use those identities, for $\alpha=x,\ \beta=\dfrac{\pi}{2}$ , you get:

$\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x$

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Hence:

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3 years ago

Question 13. Please

Sonbull [250]

Answer:

Step-by-step explanation:

Δ OPG is right with hypotenuse OG being the radius of the circle

Using Pythagoras' identity in the right triangle

OG² = OP² + PG²

[ PG = 11 , since OP is the perpendicular bisector of FG ]

OG² = 8² + 11² = 64 + 121 = 185

Δ OQS is right with hypotenuse OS being the radius

Using Pythagoras' identity

OS² = QS² + OQ² , that is

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4 = x → d

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3 years ago

A cube has a side length of x feet. Which expression represents the surface area of the cube?

alexdok [17]

The answer is D)6x2 square units.

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If side length is x, that means that a = x units.
After replacing it in the formula, you will have:
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4 years ago

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