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3 years ago

Help!! Will give brainliest and bonus air-hugs!!

Mathematics

1 answer:

natali 33 [55]3 years ago

4 0

Answer:

Step-by-step explanation:

A) either radius or diameter and height

V = πR²h or πD²h/4

B) Radius is half a diameter, R = 4 cm

C) Rather depends on how they are put together.

Placing the rectangular faces touching would make a cylinder

D) V = 3.14(4²)(2) = 100.48 ≈ 100 cm³

E) divide by 2

i.e. V = ½πR²h

F V = ½(3.14)(4²)(2) = 50.24 ≈ 50 cm³

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Which equation does an belong from <br> 15=-5 *-3<br> 4--2=6<br> 2 + -5= - 3<br> -3*-4=-12

zavuch27 [327]

Answer:

-3*-4=-12

Step-by-step explanation:

All other equations are correct. However, this equation is false, for when you multiply two negative signs, the answer is always positive:

-3 * -4 = 12

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3 years ago

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KATRIN_1 [288]

Answer:

If you multiply the amount of square inches by 5184, you'll get the number of pixels.

Step-by-step explanation:

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3 years ago

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3 years ago

Simplify the following fraction: 6/9

mamaluj [8]

2/3. you can divide the numerator and the denominator both by 3.

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3 years ago

A. Find y in terms of x if

givi [52]

Answer:

A. $y(x) =\sqrt{ 2(\frac{x^5}{5} -\frac{6^5}{5})}$

B.Therefore the solution is defined on the interval:

6≤ x ≤ +∞

Step-by-step explanation:

Given differential equation is

$\frac{dy}{dx} = x^4y^{-1}$

$\Rightarrow \frac{dy}{dx} =\frac{ x^4}{y}$

$\Rightarrow y dy = x^4dx$

Integrating both sides:

$\Rightarrow\int y dy = \int x^4dx$

$\Rightarrow \frac{y^2}{2} =\frac{x^5}{5} +C$ ........(1)

Given y(0)= 6

it means y=0 when x=6

Putting x=6 and y=0 in the equation (1)

$\therefore \frac{0^2}{2} =\frac{6^5}{5} +C$

$\Rightarrow C=-\frac{6^5}{5}$

Equation (1) become:

$\Rightarrow \frac{y^2}{2} =(\frac{x^5}{5} -\frac{6^5}{5})$

$\Rightarrow y} =\sqrt{ 2(\frac{x^5}{5} -\frac{6^5}{5})}$

Therefore $y(x) =\sqrt{ 2(\frac{x^5}{5} -\frac{6^5}{5})}$

(B)

The quantity under root must greater than or equal to zero.

Therefore,

$2(\frac{x^5}{5} -\frac{6^5}{5})} \geq0$

$\Rightarrow \frac{x^5}{5} -\frac{6^5}{5}} \geq0$

$\Rightarrow \frac{x^5}{5} \geq\frac{6^5}{5}}$

$\Rightarrow x^5\geq {6^5$

$\Rightarrow \sqrt[5]{x^5} \geq \sqrt[5]{6^5}$

$\Rightarrow x\geq {6$

When x≥0 then the value of $2(\frac{x^5}{5} -\frac{6^5}{5})}$ is real other wise the value of $2(\frac{x^5}{5} -\frac{6^5}{5})}$ is imaginary.

Therefore the solution is defined on the interval:

6≤ x ≤ +∞

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4 years ago