Can someone help me with a basic geometry question :c

Mathematics

1 answer:

viktelen [127]2 years ago

7 0

480 I’m pretty sure ( a=LxWxH)

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Who wants free gram.m.arl.y pre.mi.um acco.unt

Damm [24]

Don’t fall for this it’s fake

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3 years ago

Which of these choices show a pair of equivalent expressions? A.7^5/7 and (√ 7)^5 B.6^7/2 and (√ 6)^7 C.(4√ 81)^7 and 81^7/4 D.5

OverLord2011 [107]

We will use the following law of indices (or 'index law') to check each pair of expression

$x^{ \frac{m}{n}} = ( \sqrt[n]{x} )^{m}$

With fractional power, the denominator is the root and the numerator is the power of the term. When the denominator is 2, we usually only write the normal square symbol (√). Denominator other than 2, we usually write the value of the root, for example, the cubic root ∛

Option A - Incorrect
$7^{ \frac{5}{7}}$ should equal to $( \sqrt[7]{7} )^{5}$

Option B - Correct
$6^{ \frac{7}{2} }$ does equal to $( \sqrt{6}) ^{7}$

Option C - Incorrect
$4( \sqrt{81} )^{7}$ should equal to $(4)( 81^{ \frac{7}{2} } )$

Option D - Incorrect
$5^{ \frac{2}{3} }$ should equal to $( \sqrt[3]{5} )^{2}$

7 0

3 years ago

Read 2 more answers

Find the area of the circle x^2+y^2=16 by the method of intregration

zhuklara [117]

Answer:

Hello,

$16\pi$

Step-by-step explanation:

$I=\dfrac{Area}{4} =\int\limits^4_0 {\sqrt{16-x^2} } \, dx \\\\Let\ say\ x=4*sin(t),\ dx=4*cos(t) dt\\\\\displaystyle I=\int\limits^\frac{\pi }{2} _0 {4*\sqrt{1-sin^2(t)} }*4*cos(t) \, dt \\\\=16*\int\limits^\frac{\pi }{2} _0 {cos^2(t)} \, dt \\\\=16*\int\limits^\frac{\pi }{2} _0 {\frac{1-cos(2t)}{2}} \, dt \\\\=8*[t]^\frac{\pi }{2} _0-[\frac{sin(2t)}{2} ]^\frac{\pi }{2} _0\\\\=4\pi -0\\\\=4\pi\\\\\boxed{Area=4*I=16\pi}\\$