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umka21 [38]
3 years ago
14

Write four numbers that could be ordered between these numbers 59 through 1,000

Mathematics
2 answers:
Blababa [14]3 years ago
7 0
100, 200, 300, and 400.
Mrrafil [7]3 years ago
3 0
99, 199, 299, 399,
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4th term of (4x-y)^9
Neko [114]
Now, let's do the same as we did for the previous one here.

\bf (4x-y)^9\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&+1&(4x)^9(-y)^0\\
2&+9&(4x)^8(-y)^1\\
3&+36&(4x)^7(-y)^2\\
4&+84&(4x)^6(-y)^3
\end{array}

notice again, how did we get 84 for the 4th element's coefficient? well 36 * 7 / 3.  and so on.  And you can just expand it from there.
3 0
3 years ago
Item 7
Mariulka [41]

Answer:

A = 74.7^\circ

B = 42.5^\circ

C = 62.8^\circ

Step-by-step explanation:

Given

A = (-1,2) \to (x_1,y_1)

B = (2,8) \to (x_2,y_2)

C = (4,1) \to (x_3,y_3)

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2

For AB

A = (-1,2) \to (x_1,y_1)

B = (2,8) \to (x_2,y_2)

d = \sqrt{(-1 - 2)^2 + (2 - 8)^2

d = \sqrt{(-3)^2 + (-6)^2

d = \sqrt{45

So:

AB = \sqrt{45

For BC

B = (2,8) \to (x_2,y_2)

C = (4,1) \to (x_3,y_3)

BC = \sqrt{(2 - 4)^2 + (8 - 1)^2

BC = \sqrt{(-2)^2 + (7)^2

BC = \sqrt{53

For AC

A = (-1,2) \to (x_1,y_1)

C = (4,1) \to (x_3,y_3)

AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2

AC = \sqrt{(-5)^2 + (1)^2

AC = \sqrt{26

So, we have:

AB = \sqrt{45

BC = \sqrt{53

AC = \sqrt{26

By representation

AB \to c

BC \to a

AC \to b

So, we have:

a = \sqrt{53

b = \sqrt{26

c = \sqrt{45

By cosine laws, the angles are calculated using:

a^2 = b^2 + c^2 -2bc \cos A

b^2 = a^2 + c^2 -2ac \cos B

c^2 = a^2 + b^2 -2ab\ cos C

a^2 = b^2 + c^2 -2bc \cos A

(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A

53 = 26 +45 - 2 * 34.21 * \cos A

53 = 26 +45 - 68.42 * \cos A

Collect like terms

53 - 26 -45 = - 68.42 * \cos A

-18 = - 68.42 * \cos A

Solve for \cos A

\cos A =\frac{-18}{-68.42}

\cos A =0.2631

Take arc cos of both sides

A =\cos^{-1}(0.2631)

A = 74.7^\circ

b^2 = a^2 + c^2 -2ac \cos B

(\sqrt{26})^2 = (\sqrt{53})^2 +(\sqrt{45})^2 - 2 * (\sqrt{53}) +(\sqrt{45}) * \cos B

26 = 53 +45 -97.67 * \cos B

Collect like terms

26 - 53 -45= -97.67 * \cos B

-72= -97.67 * \cos B

Solve for \cos B

\cos B = \frac{-72}{-97.67}

\cos B = 0.7372

Take arc cos of both sides

B = \cos^{-1}(0.7372)

B = 42.5^\circ

For the third angle, we use:

A + B + C = 180 --- angles in a triangle

Make C the subject

C = 180 - A -B

C = 180 - 74.7 -42.5

C = 62.8^\circ

8 0
2 years ago
Which angle is congruent to
kondor19780726 [428]

The angle congruent to ∠CGD is the angle ∠AGF as they are vertical angles.

5 0
1 year ago
In how many places did the japanese troops land on december 8, 1941?
bezimeni [28]
The first battle in the Pacific between the empire of Japan against the US and its allies took place on December 8, 1941. The Japanese forces began their conquest on the pacific islands and it started its invasion of the Malaya. They were held by a resistance of British Indian fighter however it was not enough to hold off the overwhelming Japanese force.
7 0
3 years ago
Find the area of this triangle.
Oxana [17]

Answer:

The area is 104

Step-by-step explanation:

8 0
2 years ago
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