Solve 2(1 - x) > 2x.a.) O x < 0.5b.) Ox> 2c.) Ox<2d.) O x > 0.5

Mathematics

2 answers:

SpyIntel [72]2 years ago

7 0

Answer:

a.)x < 0.5

Step-by-step explanation:

2(1 - x) > 2x

2 - 2x > 2x

+2x +2x

2 > 4x

/4 /4

0.5 > x

barxatty [35]2 years ago

4 0

Answer:

a) Ox<0.5 wjsjdhdjdjdj

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Angle 0 lies in the second quadrant and sin 0 = 3/5 cos= coy =

Elza [17]

Consider the right triangle ABC with legs AB=4, AC=3 and hypotenuse BC=5. Angle B has
$\sin B=\frac{opposite}{hypotenuse} = \frac{3}{5}$ and
$\cos B = \frac{adjacent}{hypotenuse} = \frac{4}{5}$ .

Since O lies in second quadrant $\cos O\ \textless \ 0$ and
$\cos O= -\cos B =- \frac{4}{5}$ .
Answer: $\cos O=- \frac{4}{5}$ .

3 0

3 years ago

Please Help!

alisha [4.7K]

Diagonal of the parallelogram divides the parallelogram in to two equal areas.

So area of parallelogram = 2(area of triangle)

According to the given diagram,

AB= 8, AD = 5 and BD = 11

So according to the Heron's formula,

Area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}\\\\where,\\ s =\frac{a + b + c }{2}$

and a, b and c are the three sides of the triangle

Area of triangle ABD = $s=\frac{8 + 5 + 11}{2} \\\\s = \frac{24}{2}\\\\s = 12Area of triangle ABD = \sqrt{12(12 - 8) (12 - 5) (12 - 11)}\\\\Area of triangle ABD = \sqrt{12(4) (7) (1)}\\\\Area of triangle ABD = \sqrt{336}\\\\Area of triangle ABD = 18.33$