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trapecia [35]
3 years ago
9

A

Mathematics
1 answer:
ozzi3 years ago
3 0

Answer:

m<DEF = 47°

Step-by-step explanation:

Given:

m(DEF) = 266°

Required:

m<DEF

Solution:

m<DCF = ½(m(DEF)) (Inscribed angle theorem)

Substitute

m<DCF = ½(266°)

m<DCF = 133°

Opposite angles in an inscribed quadrilateral are supplementary.

Therefore,

m<DEF = 180° - m<DCF

m<DEF = 180° - 133°

m<DEF = 47°

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Y= 5x -3 what does y=
irga5000 [103]

Answer:

15

Step-by-step explanation:

6 0
3 years ago
Which of the following inequalities is best represented by this graph?
Nataly [62]
For this case, the first thing to do is find the equation of the line correctly and then see which of the inequalities represents the shaded region.
 We have then:
 5x - y = 2
 For x = 0 
 5 (0) - y = 2
 -y = 2
 y = -2
 For y = 0
 5x - 0 = 2
 5x = 2
 x = 2/5 = 0.4
 Therefore the line is:
 5x - y = 2 
 Then, we see that for the point (0, 0)
 5x - y ≥ 2
 5 (0) - (0) ≥ 2
 0 ≥ 2
 The inequality is not met, which is correct because this point is not part of the shaded region.
 Answer:
 The inequality that best is represented by the graph if: 
 5x - y ≥ 2
5 0
3 years ago
If sin square theta = 0.75, then cos =
sergeinik [125]

Answer:

\cos\theta=\pm0.5

Step-by-step explanation:

Given:

\sin^2\theta=0.75

To find \cos\theta

Using trigonometric relations for sums and differences of squares of the ratios.

We know:

\sin^2\theta+\cos^2\theta =1

Plugging in  \sin^2\theta=0.75 in the above relation.

0.75+\cos^2\theta =1

Subtracting both sides by 0.75.

0.75+\cos^2\theta-0.75 =1-0.75

\cos^2\theta =0.25

Taking square root both sides.

\sqrt{\cos^2\theta} =\sqrt{0.25}

∴ \cos\theta=\pm0.5  (Answer)

6 0
3 years ago
What is the value of y=4x−2 when x=3? When x=3, y=
bonufazy [111]
The answer to you question is y=10
4 0
3 years ago
*****PLEASE HELP (30 POINTS)*****
ziro4ka [17]

Answer:

2. C) 21.2 yd    3. B) angle IQL    4. B) 55 degree

Step-by-step explanation:

2. The area of an equilateral triangle is given by,

A = \frac {\sqrt{3} \times a^{2}}{4}------(1)  where a = length of each side.

Here, a = 7 yd

so , putting a = 7 yd in (1) , we get,

A = \frac {\sqrt{3} \times 49}{4}  sq. yd.

  \simeq 21.2 sq. yd.

3. B) angle IQL

4. Let, the centre of the circle be at O,

then, ∠DOB = 105°  and ∠BOC = 145° (given)

so, ∠OBD = \frac{180^{\circ} - \angle DOB}{2}

                = \frac{180^{\circ} - 105^{\circ}}{2}

                = 37.5°  [since the sum of three angles of a triangle  is 180° and OB and OD are two radii of the same circle]

and, similarly, ∠OBC = \frac{180^{\circ} - \angle COB}{2}

                = \frac{180^{\circ} - 145^{\circ}}{2}

                = 17.5° [since the sum of three angles of a triangle  is 180° and OB and OC are two radii of the same circle]

so,  ∠CBD = ∠OBD + ∠OBC

                 = 37.5° + 17.5°

                 = 55°

3 0
3 years ago
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