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3 years ago

6

Which of these could be the length of sides of right triangle ABC?

1 answer:

Anna007 [38]3 years ago

4 0

Answer:

C

Step-by-step explanation:

5*5+12*12=13*13

25+144=169

169=169

Hope this helps you!

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1 point (1) Three times a number minus two times a number* Your answer<br>

jekas [21]

Answer:

Three times a number minus two times a number

3x-2x

=x

7 0

2 years ago

PLSSSSSS ANGLES I need helppppp

kirill115 [55]

A because they are parallel and equal

4 0

1 year ago

Read 2 more answers

Find the area of the regular polygon

olasank [31]

$area = 44.0467 \: cm {}^{2}$

<h3>step by step explanation </h3>

● area of regular polygon Formulas ,

⟹ $a = \frac{l^{2}n}{4tan(\frac{\pi }{n)}}$

● now finding the area of regular polygon,

⟹ $\frac{4 {}^{2} \times 8 }{ 4 \: tan \: \binom{180}{5} }$

⟹ $a = \frac{128}{4 \times 0.7265}$

⟹ $a = 44.0467 \: cm {}^{2}$

Hope it's helps you

4 0

2 years ago

What is the area of this composite figure?

11Alexandr11 [23.1K]

Answer:

256 cm squared

Step-by-step explanation:

Just did the Quiz, hope this helps!

4 0

3 years ago

Read 2 more answers

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

2 years ago