$\sqrt[3]{275:7}=\sqrt[3]{\dfrac{275}{7}}=\dfrac{\sqrt[3]{275}}{\sqrt[3]{7}}=\dfrac{\sqrt[3]{275}\cdot\sqrt[3]{7^2}}{\sqrt[3]{7}\cdot\sqrt[3]{7^2}}=\dfrac{\sqrt[3]{275\cdot49}}{\sqrt[3]{7\cdot7^2}}\\\\=\dfrac{\sqrt[3]{13475}}{\sqrt[3]{7^3}}=\dfrac{\sqrt[3]{13475}}{7}=\dfrac{1}{7}\sqrt[3]{13475}$

3 0

2 years ago

If Acot theta + Bcot theta =4 and

geniusboy [140]

Answer is: B

Step-by-step explanation: correct answer is B

−a

Given, acotθ+bcscθ=p

bcotθ+acscθ=q

⇒p

−q

=(p−q)(p+q)

=[acotθ+bcscθ−bcotθ−acscθ][acotθ+bcscθ+bcotθ+acscθ]

=[cotθ(a−b)−cscθ(a−b)][cotθ(a+b)+cscθ(a+b)]

=(a−b)(cotθ−cscθ)(a+b)(cotθ+cscθ)

=(a

−b

)(cot

θ−csc

θ)

=(−1)(a

−b

)[∵csc

θ−cot

θ=1]

=(b

−a

)

7 0

2 years ago

Can I get some help on this one and if you could show its done I would appreciate it. Thank you.

Andreyy89

<h3>Answer: 5.5 which is choice B</h3>

Have a look at the diagram I posted below. I marked on your image to add in another angle. This angle is also 20 degrees because of congruent alternate interior angles (horizontal lines are parallel). This angle I add in is the reference angle of the triangle

opposite the reference angle is the vertical side x

adjacent to the reference angle is the horizontal side 15

We'll use the tangent rule. Make sure your calculator is in degree mode.

tan(angle) = opposite/adjacent

tan(20) = x/15

15*tan(20) = x <<-- multiply both sides by 15

x = 15*tan(20)

x = 5.45955 <<--- use calculator; this is approximate

x = 5.5 <<--- round to one decimal place

5 0

3 years ago