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alex41 [277]
3 years ago
7

The minute hand of a clock is exactly 6in in length, and the tip of the minute hand has traveled ten inches since noon. What tim

e is it?
Mathematics
1 answer:
frutty [35]3 years ago
3 0
\bf \textit{arc's length}\\\\
s=\cfrac{\theta \pi r}{180}\qquad 
\begin{cases}
r=radius\\
\theta =angle\ in\\
\qquad degrees\\
------\\
r=6\\
s=10
\end{cases}\implies 10=\cfrac{\theta \pi 6}{180}\implies \cfrac{180\cdot 10}{6\pi }=\theta 
\\\\\\
\cfrac{300}{\pi }=\theta \implies 95.49^o\approx \theta

now, the circle of the clock has 360°, if we divide it by 60(minutes), we get 360/60, just 6° for each minute.

now, if there are 6° in 1 minute, how many minutes in 95.49°?

well, just 95.49/6 or about 15.92 minutes, I take it you can round it up to 16 minutes.

so 16 minutes since noon, so is about 12:16, about time get the silverware for lunch.
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Answer:

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Step-by-step explanation:

Container given in the picture is in the shape of a cuboid.

And volume of a cuboid is measured by the expression,

Volume of a cuboid = Length × width × height

Now substitute the measure of the container's dimensions given in the picture

Volume = (4x² + 3x)(x²- 8)(6x + 15)

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             = [4x²(x² - 8) + 3x(x² - 8)](6x + 15)

             = (4x⁴ - 32x² + 3x³ - 24x)(6x + 15)

             = (4x⁴ + 3x³ - 32x² - 24x)(6x + 15)

             = 6x(4x⁴ + 3x³ - 32x² - 24x) + 15(4x⁴ + 3x³ - 32x² - 24x)

             = 24x⁵+ 18x⁴ - 192x³ - 144x² + 60x⁴ + 45x³ - 480x² - 360x

             = 24x⁵ + 78x⁴ - 147x³ - 624x² - 360x

8 0
3 years ago
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irina1246 [14]

ans is b...multiply eqn 1 by -2 and eqn 2 by 3

=》 -6x + 10y = 4...eqn 3

=》 6x + 3y = 9...eqn 4

add eqn 3 and 4...

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=》y =1 and x = 1

5 0
3 years ago
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natta225 [31]

Answer:

UV=29

Step-by-step explanation:

In right triangles AQB and AVB,

∠AQB = ∠AVB ...(i)  {Right angles}

∠QBA = ∠VBA ...(ii)  {Given that they are equal}

We know that sum of all three angles in a triangle is equal to 180 degree. So wee can write sum equation for each triangle


∠AQB+∠QBA+∠BAQ=180 ...(iii)

∠AVB+∠VBA+∠BAV=180 ...(iv)


using (iii) and (iv)

∠AQB+∠QBA+∠BAQ=∠AVB+∠VBA+∠BAV

∠AVB+∠VBA+∠BAQ=∠AVB+∠VBA+∠BAV  (using (i) and (ii))

∠BAQ=∠BAV...(v)


Now consider triangles AQB and AVB;

∠BAQ=∠BAV  {from (v)}

∠QBA = ∠VBA {from (ii)}

AB=AB  {common side}

So using ASA, triangles AQB and AVB are congruent.

We know that corresponding sides of congruent triangles are equal.

Hence

AQ=AV

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7 0
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Illusion [34]

we know that


step 1

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so

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step 2

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