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3 years ago

Question no. e. and f. please help me !!!!!

Mathematics

2 answers:

erastovalidia [21]3 years ago

8 0

If you need help again get that app photomath it is amazing(not sponsered)

Verizon [17]3 years ago

4 0

Answer: see proofs below

Step-by-step explanation:

Proofs are like a puzzle that you have to unravel.

You can manipulate the left side to get the right side (LHS → RHS)

You can manipulate the right side to get the left side (RHS → LHS)

You will need to use Pythagorean Identities, Sum and Difference Identities, and/or Double Angle Identities when manipulating one of the sides.

Use the following Identities:

cos 2A = cos²A - sin²A

sin 2A = 2 cosA · sinA Note that A can be substituted with 2A

sin 3A = 3 sinA - 4sin³A

cos 3A = 4cos³A - 3cosA

***********************************************************************************

e) Prove: 4 sin A · cos³A - 4 cos A · sin³A = sin(4A)

LHS → RHS

Given: 4 sin A · cos³A - 4 cos A · sin³A

Factor: 4 sin A · cos x (cos²A - sin²A)

2(2 sin A · cos A)(cos²A - sin²A)

Double Angle Identity: 2(sin 2·A)(cos 2·A)

2 sin (2A) · cos (2A)

Double Angle Identity: sin (2·2A)

sin 4A

Proven: sin 4A = sin 4A $\checkmark$

*****************************************************************************************

f) Prove: 4 sin³A · cos 3A + 4 cos³A · sin 3A = 3 sin(4A)

LHS → RHS

Given: 4 sin³A · cos 3A + 4 cos³A · sin 3A

Triple Angle Identity: 4sin³A (4cos³A - 3 cosA) + 4cos³A (3sinA - 4sin³A)

Expand: 16sin³A · cos³A - 12sin³A · cosA + 12cos³A · sinA - 16sin³A · cos³A

Simplify: 12cos³A · sinA - 12sin³A · cosA

Expand: 6cos²A(2cosA · sinA) - 6sin²A(2cosA · sinA)

Double Angle Identity: 6cos²A(sin 2A) - 6sin²A(sin 2A)

Factor: 6sin 2A (cos²A - sin²A)

Double Angle Identity: 6sin 2A (cos 2A)

3(2sin 2A · cos 2A)

Simplify: 3sin (2·2A)

3 sin 4A

Proven: 3sin 4A = 3sin4A $\checkmark$

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zavuch27 [327]

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3 years ago

Three of the angles of a

irina1246 [14]

Answer:

Angles of a quadrilateral add up to 360

so x+36+72+144=360

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So B is the correct answer

Step-by-step explanation:

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3 years ago

Quadrilateral ABCDABCD has vertices AA (-2, 2), BB (4, 5), CC (3, 0) and DD (-3, -3). What are the coordinates of the midpoint o

Arada [10]

Check the picture below.

the quadrilateral is a parallelogram, thus the diagonals bisect each other, so, we could have used either diagonal, and get the same midpoint.

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
B&({{ 4}}\quad ,&{{ 5}})\quad 
% (c,d)
D&({{ -3}}\quad ,&{{ -3}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
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4 years ago

The hypotenuse of a right triangle measures 15 cm and one of its legs

Kay [80]

Answer:

[see below]

Step-by-step explanation:

Use the Pythagorean Theorem.

$a^2+b^2=c^2$

'c' will be the hypotenuse.

The hypotenuse is 15 cm. One of the legs is 5 cm.

$5^2+b^2=15^2\\\\25 + b^2 = 225\\\\25-25+b^2=225-25\\\\b^2=200\\\\\sqrt{b^2}=\sqrt{200}\\\\ b=14.1421356237..., -14.1421356237$

14.1421356237 ≈ 14.1

The other leg should be about 14.1 cm.

Hope this helps.

4 0

3 years ago

Read 2 more answers

What is the product? (-2d^2+s)(5d^2-6s)

Anna71 [15]

Answer:

Option 1: -10d⁴ + 17d²s - 6s²

General Formulas and Concepts:

Algebra I

Terms/Coefficients
Expand by FOIL

Step-by-step explanation:

Step 1: Define

Identify

(-2d² + s)(5d² - 6s)

Step 2: Expand

FOIL: -10d⁴ + 12d²s + 5d²s - 6s²
Combine like terms: -10d⁴ + 17d²s - 6s²

6 0

3 years ago

Question no. e. and f. please help me !!!!!​

Question no. e. and f. please help me !!!!!