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3 years ago

Pls help me.i need help asap

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

3 0

Answer:

Area of the shaded region = 51 cm²

Step-by-step explanation:

Area of the smaller pentagon ABKHG = 17 cm²

Since, B is the midpoint of side AC,

Scale factor used to enlarge the sides = $\frac{\text{Measure of side AC}}{\text{Measure of side AB}}$

= $\frac{AC}{\frac{AC}{2}}$

= 2

Therefore, area scale factor of both the pentagons = (Dimensional scale factor)²

2² = $\frac{\text{Area of AFEDC}}{\text{Area of ABKHG}}$

4 = $\frac{\text{Area of AFEDC}}{17}$

Area of bigger pentagon AFEDC = 4 × 17

= 68 cm²

Area of the shaded region = Area of the bigger pentagon - Area of the smaller pentagon

= 68 - 17

= 51 cm²

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artcher [175]

Answer:

The root of x = -2 + √2

Step-by-step explanation:

Step(i):-

Given $\sqrt{2x^{2} } + x +\sqrt{2} = 0$

⇒ $\sqrt{2} } ( x^2 ) ^{\frac{1}{2} } + x +\sqrt{2} = 0$

⇒ $\sqrt{2}x + x +\sqrt{2} = 0$

⇒ $(\sqrt{2} + 1) x = - \sqrt{2}$

$x = \frac{-\sqrt{2} }{(\sqrt{2} + 1)}$

Step(ii):-

Rationalizing

$x = \frac{-\sqrt{2} }{(\sqrt{2} + 1)} X \frac{(\sqrt{2} - 1)}{(\sqrt{2} - 1)}$

Apply ( a-b ) ( a+b) = a²-b²

$x = \frac{-\sqrt{2}(\sqrt{2} -1) }{(\sqrt{2})^{2} - 1^{2} )} = \frac{-(2-\sqrt{2} }{2-1}$

The root of x = -2 + √2

6 0

3 years ago

How can you split 20 into 3 groups

Leviafan [203]

The answer is 6 per team with 2 left over, as (6 x 3) + 2 = 20

I thought of the largest amount that could be multiplied by 3, but the product would be less or equal to 20, that's how I got 6.

Then to find the left over remaining balls that are given out I subtracted 20-18 = 2.

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3 years ago

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Alexus [3.1K]

Answer: D) 1.601 x 10 ^9

Step-by-step explanation:

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3 years ago

An engineer, in an attempt to make a quick measurement, walks 100 ft from the base of an overpass

Korvikt [17]

Answer:

The distance of the overpass above the ground is approximately 26.795 ft

Step-by-step explanation:

The parameters given are;

The distance from the overpass the engineer stands before determining the angle of elevation of the overpass from his standing point = 100 ft

The angle of elevation of the overpass as determined by the engineer from 100 ft = 15°

By trigonometric ratios, we have;

$Tan(\theta) = \dfrac{Opposite \, side \, to\ angle}{Adjacent\, side \, to\, angle}$

The opposite side to the 15° angle of elevation in the above case is the distance of the overpass above the ground

The opposite side to the 15° is the distance of the engineer from the base of the overpass

Therefore;

Tan(15°) the height of the overpass=

length

$Tan(15 ^{\circ}) = \dfrac{The \ distance \, of \, the \ overpass \ above \ ground}{100 \ ft}$

The distance of the overpass above the ground = 100 × tan (15°) ≈ 26.795 ft.

6 0

3 years ago

Look at photo. plz help. need it soon. will mark brainliest

Mandarinka [93]

Answer:
Both Ivan and Polina is correct.

5 0

3 years ago

Pls help me.i need help asap​

Pls help me.i need help asap