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4 years ago

10

Help (pic attached!!)

1 answer:

mash [69]4 years ago

7 0

$( \sqrt[5]{ 3^{2} } )^{ \frac{1}{3} } = (3^{ \frac{2}{5} } ) ^{ \frac{1}{3} } =$ $3^{ \frac{2}{15} }$

Answer: C.

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What is the developed form of 2(3x-10)

vovikov84 [41]

The first step to finding the developed form is to multiply each term in the parenthesis by 2
2 × 3x - 2 × 10
now,, youll need to calculate the product of the first multiplication set
6x - 2 × 10
finally,, multiply the last set of numbers
6x - 20
this means that the correct answer to your question is 6x - 20.
let me know if you have any further questions
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4 years ago

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Rearranging formulae:

bazaltina [42]

Answer:

1. d/a+c=d

2. (m+21)/5=n

3. (1/2+2q)*4=p or 2+8q=p

4. (p-2a)/2pi=r

5. {[(5c+1)/2]+c}/3=a

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3 years ago

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What is the vertex of the parabola given by = -(x-2)^2-1?

weqwewe [10]

Rewrite the parabola in standard form and use this formula to find the vertex (h,k).
= (2,-1)

3 0

3 years ago

At a restaurant the bill was $35.10 with a 15% tip added on. How much is the total bill, rounded to the nearest whole cent?

blondinia [14]

Answer:

15% = 0.15

35.10 * (1 + 0.15) = $40.37

7 0

3 years ago

A vertical cylinder is leaking water at a rate of 4m3/sec. If the cylinder has a height of 10m and a radius of 2m, at what rate

Lyrx [107]

Answer:

Therefore the rate change of height is $\frac{1}{\pi}$ m/s.

Step-by-step explanation:

Given that a vertical cylinder is leaking water at rate of 4 m³/s.

It means the rate change of volume is 4 m³/s.

$\frac{dV}{dt}=4 \ m^3/s$

The radius of the cylinder remains constant with respect to time, but the height of the water label changes with respect to time.

The height of the cylinder be h(say).

The volume of a cylinder is $V=\pi r^2 h$

$=( \pi \times 2^2\times h)\ m^3$

$\therefore V= 4\pi h$

Differentiating with respect to t.

$\frac{dV}{dt}=4\pi \frac{dh}{dt}$

Putting the value $\frac{dV}{dt}$

$\Rightarrow 4\pi \frac{dh}{dt}=4$

$\Rightarrow \frac{dh}{dt}=\frac{4}{4\pi}$

$\Rightarrow \frac{dh}{dt}=\frac{1}{\pi}$

The rate change of height does not depend on the height.

Therefore the rate change of height is $\frac{1}{\pi}$ m/s.

3 0

3 years ago