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3 years ago

The volume of a cube is 9000 m3 .if the volume of the cube was measured in cm3 ,which would change?

Mathematics

1 answer:

luda_lava [24]3 years ago

8 0

volume of cube is measured as 900000 cubic centimetre

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What is the sum of the arithmetic series 18 t=1 (3t-4)

PolarNik [594]

Consider the arithmetic series $_{t=1}\Sigma^{t=18} (3t-4)$

Let t=1 in the given series, we get

first term = $a_{1}$ = 3-4 = -1.

Let t=2 in the given series, we get

second term = $a_{2}$ = $(3 \times 2)-4 = 2$

Let t=3 in the given series, we get

third term = $a_{3} = (3 \times 3)-4=5$

Now, let t=18 in the given series, we get

last term = l = $l = (3 \times 18)-4 = 50$

We get the series as

-1, 2, 5,..... 50

Sum = $\frac{n}{2}(a+l)$

= $\frac{18}{2}(-1+50)$

= $= 9 \times 49$

= 441

Therefore, the sum of the given arithmetic series is 441.

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4 years ago

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Help me on this question with a explanation please.

liq [111]

Answer is 50cm, to get the diameter you divide the circumference by pi.

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3 years ago

Reason for 4 pleaseeee

Marina CMI [18]

The reason for statement 4 is CPCTC.

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3 years ago

Use the squared identities to simplify 2sin^2xcos^2x

Rainbow [258]

$\bf sin^2({{ \theta}})=\cfrac{1-cos(2{{ \theta}})}{2}\qquad \qquad 

cos^2({{ \theta}})=\cfrac{1+cos(2{{ \theta}})}{2}\\
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thus
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\\\\
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\cfrac{[1-cos(2x)][1+cos(2x)]}{2}\\
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\\ \quad \\
$

$\bf (a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\
-----------------------------\\
thus
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\cfrac{[1-cos(2x)][1+cos(2x)]}{2}\implies \cfrac{1^2-cos^2(x)}{2}
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\begin{array}{llll}
\textit{pretty sure you know what that numerator is}\\
\textit{check your pythagorean identities}
\end{array}$

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3 years ago

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Factor. x2+14x+48 Enter your answer in the box.

Softa [21]

Answer:

(x+6)(x+8)

Step by step explanation:

Because 6 + 8 is the 14 in the middle and 48 is 6 x 8 and you thats all you need just put the x's at the beginning

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