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3 years ago

9

A

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