Answer:
your answer should be number 1
<h2>9.</h2><h3>Given</h3>
<h3>Find</h3>
- linear approximation to the volume when the radius increases 0.4 cm
<h3>Solution</h3>
The equation for volume of a sphere is
... V = (4/3)π·r³
Differentiating gives
... dV = 4π·r²·dr
Filling in the given numbers gives
... change in volume ≈ 4π·(15 cm)²·(0.4 cm)
... = 360π cm³ ≈ 1130.97 cm³ . . . . . . volume of layer 4mm thick
<h2>11.</h2><h3>Given</h3>
- an x by x by 2x cuboid with surface area 129.6 cm²
- rate of change of x is 0.01 cm/s
<h3>Find</h3>
<h3>Solution</h3>
The area is that of two cubes of dimension x joined together. The area of each such cube is 6x², but the two joined faces don't count in the external surface area. Thus the area of the cuboid is 10x².
The volume of the cuboid is that of two cubes joined, so is 2x³. Then the rate of change of volume is
... dV/dt = (d/dt)(2x³) = 6x²·dx/dt
We know x² = A/10, where A is the area of the cuboid, so the rate of change of volume is ...
... dV/dt = (6/10)A·dx/dt = 0.6·(129.6 cm²)(0.01 cm/s)
... dV/dt = 0.7776 cm³/s
The value of x is 1/13 (4th choice)
Step-by-step explanation:
![{a}^{ \frac{m}{n} } = \sqrt[n]{ {a}^{m} }](https://tex.z-dn.net/?f=%20%7Ba%7D%5E%7B%20%5Cfrac%7Bm%7D%7Bn%7D%20%7D%20%20%3D%20%20%5Csqrt%5Bn%5D%7B%20%7Ba%7D%5E%7Bm%7D%20%7D%20)
If 14^x = ¹³√14, what value of <em>x </em>makes this equation true?
![\sqrt[13]{14} = {14}^{x} \\ \sqrt[13]{14} = {14}^{ \frac{1}{13} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B13%5D%7B14%7D%20%20%3D%20%20%7B14%7D%5E%7Bx%7D%20%20%5C%5C%20%20%5Csqrt%5B13%5D%7B14%7D%20%20%3D%20%20%7B14%7D%5E%7B%20%5Cfrac%7B1%7D%7B13%7D%20%7D%20)
So, the value of <em>x</em> is 1/13 (4th choice)
<em>Hope</em><em> </em><em>it </em><em>helpful </em><em>and </em><em>useful </em><em>:</em><em>)</em>
Answer:
7.416198
Step-by-step explanation:
You must find out which number produces 55 when multiplied by itself
