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3 years ago

What’s the answer??? ( ONLY ANSWER IF YOU KNOW)

Mathematics

1 answer:

Elena-2011 [213]3 years ago

4 0

The exponents of expression one are the same as the exponents of expression two

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A painter takes 2 3/5 hours to paint a wall. How many hours will the painter take to paint 8 walls if she works at the same rate

Naya [18.7K]

2 3/5 X 8 = 20.8 hours

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3 years ago

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Please help with this algebra question

valina [46]

Answer:

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Step-by-step explanation:

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3 years ago

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

Fofino [41]

Answer: see proof below

Step-by-step explanation:

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

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

Proof LHS → RHS

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

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

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

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

$\text{Identity:}\qquad \qquad \dfrac{\sin^2 A}{1-\cos A}$

$\text{LHS = RHS:}\quad \dfrac{\sin^2 A}{1-\cos A}=\dfrac{\sin^2 A}{1-\cos A}\quad \checkmark$

3 0

3 years ago

Given that cot θ = 1/√5, what is the value of (sec²θ - cosec²θ)/(sec²θ + cosec²θ) ?

Bogdan [553]

Step-by-step explanation:

$\mathsf{Given :\;\dfrac{{sec}^2\theta - co{sec}^2\theta}{{sec}^2\theta + co{sec}^2\theta}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{{sec}\theta = \dfrac{1}{cos\theta}}}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{co{sec}\theta = \dfrac{1}{sin\theta}}}}$

$\mathsf{\implies \dfrac{\dfrac{1}{cos^2\theta} - \dfrac{1}{sin^2\theta}}{\dfrac{1}{cos^2\theta} + \dfrac{1}{sin^2\theta}}}$

$\mathsf{\implies \dfrac{\dfrac{sin^2\theta - cos^2\theta}{sin^2\theta.cos^2\theta}}{\dfrac{sin^2\theta + cos^2\theta}{sin^2\theta.cos^2\theta}}}$

$\mathsf{\implies \dfrac{sin^2\theta - cos^2\theta}{sin^2\theta + cos^2\theta}}$

Taking sin²θ common in both numerator & denominator, We get :

$\mathsf{\implies \dfrac{sin^2\theta\left(1 - \dfrac{cos^2\theta}{sin^2\theta}\right)}{sin^2\theta\left(1 + \dfrac{cos^2\theta}{sin^2\theta}\right)}}$