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sleet_krkn [62]
3 years ago
11

Multiply: (2sin B+cos B)by 3cosec B.secB​

Mathematics
1 answer:
adoni [48]3 years ago
6 0

Step-by-step explanation:

Is this the full question

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X to the 3rd power= 64/343
leva [86]

Answer:

x = 4/7

Step-by-step explanation:

x^3 = 64/ 343 // - 64/ 343

x^3 - ( 64/ 343 ) = 0

x^3 - 64 / 343 = 0

1*x^3 = 64/ 343 // : 1

x^3 = 64/ 343

x^3 = 64/ 343 // ^ 1/3

x = 4/ 7

5 0
3 years ago
The equation Y-3=-2(x+5) is written in point-slope form. What is the y-intercept of the line?
xz_007 [3.2K]

Answer:

Step-by-step explanation:

hello :

y-3=-2(x+5)

y - intercept  when : x=0 : y-3 =-2(0+5)

y-3 =-10

y = - 7

7 0
3 years ago
Read 2 more answers
Please help, I’m not sure what the answer is.
Karo-lina-s [1.5K]

Answer:

The answer is 'C'

Step-by-step explanation:

Congruent simply means "the same." When a line divides two parralel lines transversals are created. This means that there are angles on both lines that are similar. To fully understand which angles would be the same, I reccommend researching "transversals" for more information, or asking your teacher about it as it is hard to explain without a proper diagram.

5 0
3 years ago
Could you please explain <br><br> find an equation for i, -i, -4, 1 = x as its solution
Virty [35]

We can do this easily using 0s.

(x - i) (x + i) (x + 4) (x - 1) = 0

If you plug in any of the numbers, you'll get 0, making the equation true.

5 0
3 years ago
Find the area of the right triangle △DEF with the points D (0, 0), E (1, 1), and F.
Tems11 [23]

Answer:

The area of the triangle is  \sqrt{3}

Step-by-step explanation:

Given:

Coordinates D (0, 0), E (1, 1)

Angle  ∠DEF = 60°

△DEF is a Right triangle

To Find:

The area of the triangle

Solution:

The area of the triangle is  = \frac{1}{2}(base \times height)

Here the base is Distance between D and E

calculation the distance using the distance formula, we get

DE  = \sqrt{(0-1)^2 + (0-1)^2}

DE =\sqrt{(-1) ^2 + (-1)^2

DE = \sqrt{1+1}

DE = \sqrt{2}

Base = \sqrt{2}

Height is DF

DF =tan(60^{\circ}) \times DE

DF = \sqrt{3} \times DE

DF = \sqrt{3} \times\sqrt{2}

Now, the area of the triangle is

= \frac{1}{2}({\sqrt{2})(\sqrt{3} \times \sqrt{2})

=\frac{1}{2}({\sqrt{2})(\sqrt{3} \sqrt{2})

=\frac{1}{2}(2\sqrt{3} )

=\sqrt{3}

6 0
3 years ago
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