Find three consecutive odd integers such that the sum of the first, two time the second and three times the third is 46

Mathematics

1 answer:

Andreyy893 years ago

8 0

5+(2*7)+(3*9) = 46

I'm assuming this is what you meant.

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WHAT VALUE IS EQUIVALENT TO (7+3)^2 + (8-1) X 5 EQUAL

sdas [7]

Answer:

135

Step-by-step explanation:

(7+3)^2 + (8-1) * 5

Parentheses first

(10)^2 + (7) * 5

Exponents next

100 + 7*5

Multiply

100 + 35

Add

135

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2 years ago

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3 years ago

1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

<u>Step-by-step explanation:</u>

$\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}$

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

<u>Proof LHS → RHS</u>

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$