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Karo-lina-s [1.5K]
3 years ago
10

Hi guys can u guys help me add 322 + 12

Mathematics
2 answers:
Zina [86]3 years ago
4 0

Answer:

334 is the answer

Step-by-step explanation:

you add them together

iren2701 [21]3 years ago
3 0

Answer:

334 is the right answer

Step-by-step explanation:

please mark me brainliest

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The radius of a right circular cone is increasing at a rate of 1.4 in/s while its height is decreasing at a rate of 2.1 in/s. At
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Given:

\dfrac{dr}{dt}=1.4\text{ in/s}

\dfrac{dh}{dt}=-2.1\text{ in/s}

To find:

The rate of change in volume at r=120\text{ in. and }175\text{ in.}

Solution:

We know that, volume of a cone is

V=\dfrac{1}{3}\pi r^2h

Differentiate with respect to t.

\dfrac{dV}{dt}=\dfrac{1}{3}\pi\times \left[(r^2\dfrac{dh}{dt}) + h(2r\dfrac{dr}{dt})\right]

Substitute the given values.

\dfrac{dV}{dt}=\dfrac{1}{3}\times \dfrac{22}{7}\times \left[(120)^2(-2.1) +175(2)(120)(1.4)\right]

\dfrac{dV}{dt}=\dfrac{22}{21}\times \left[-30240+58800\right]

\dfrac{dV}{dt}=\dfrac{22}{21}\times 28560

\dfrac{dV}{dt}=29920

Therefore, the volume of decreased by 29920 cubic inches per second.

6 0
3 years ago
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Answer: y=1/3x+4

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maksim [4K]

Answer:

The value of the first "5" in the number 255,\!120 is ten times that of the second "5\!" in this number.

Step-by-step explanation:

What gives the number "255,\!120" its value? Of course, each of its six digits has contributed. However, their significance are not exactly the same. For example, changing the first \verb!5! to \verb!6! would give 2\mathbf{6}5,\!120 and increase the value of this number by 10,\!000. On the other hand, changing the second \verb!5!\! to \verb!6!\! would give 25\mathbf{6},\!120, which is an increase of only 1,\!000 compared to the original number.

The order of these two digits matter because the number "255,\!120" is written using positional notation. In this notation, the position of each digits gives the digit a unique weight. For example, in 255,\!120\!:

\begin{array}{|r||c|c|c|c|c|c|}\cline{1-7}\verb!Digit!& \verb!2! & \verb!5! & \verb!5! & \verb!1! & \verb!2! & \verb!0!\\\cline{1-7}\textsf{Index} & 5 & 4 & 3 & 2 & 1& 0 \\ \cline{1-7} \textsf{Weight} & 10^{5} & 10^{4} & 10^{3} & 10^{2} & 10^{1} & 10^{0}\\\cline{1-7}\end{array}.

(Note that the index starts at 0 from the right-hand side.)

Using these weights, the value 255,\!120 can be written as the sum:

\begin{aligned}& 255,\!120\\ &= 2 \times 10^{5} + 5 \times 10^{4} + 5 \times 10^{3} + 1 \times 10^{2} + 2 \times 10^{1} + 0 \times 10^{0} \\&=200,\!000 + 50,\!000 + 5,\!000 + 100 + 20 + 0 \end{aligned}.

As seen in this sum, the first "5" contributed 50,\!000 to the total value, while the second "5\!" contributed only 5,\!000.

Hence: The value of the first "5" in the number 255,\!120 is ten times that of the second "5\!" in this number.  

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Make them have the same denominator.

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