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2 years ago

Fide the volume and round to the nearest tenth . Please help I will give you brainless

Mathematics

2 answers:

devlian [24]2 years ago

5 0

Answer:

Step-by-step explanation:

Hatshy [7]2 years ago

3 0

Answer:

4.18 inch^3

Step-by-step explanation:

use the sphere volume formula :

V = (4/3)*pi*r^3

the radius of this sphere is 2/2 = 1 ( diameter / 2)

plugging in the values :

V = (4/3)*3.1415*1^3 = 4.18

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A solid oblique pyramid has a regular hexagon base with an area of 54_/3 cm^2 and an edge length of 6cm. Angle BAC measures 60 d

Lady_Fox [76]

Answer: The answer is (C) 324 cubic cm.

Step-by-step explanation: As given in the question, given a solid oblique pyramid with a regular hexagonal base and area 54√3 cm². Also, the edge length of the base is 6cm and ∠BAC = 60°.

We are to find the volume of the pyramid.

The formula for finding the volume of a pyramid is given by

$V=\dfrac{1}{3}b\times h,$

where, 'b' is the base area and 'h' is the perpendicular height of the pyramid.

Here, b = 54√3 cm², h = ?

Now, from the right-angled triangle ABC, we have

$\dfrac{\textup{BC}}{\textup{AC}}=\tan 60^\circ\\\\\Rightarrow \dfrac{h}{6}=\sqrt 3\\\\\Rightarrow h=6\sqrt 3.$

Therefore, the volume of the pyramid is

$V=\dfrac{1}{3}b\times h=\dfrac{1}{3}\times 54\sqrt 3\times 6\sqrt 3=324.$

Thus, the required volume is $V=324~\textup{cm}^3.$ This makes (C) as the correct option.

5 0

3 years ago

Over the course of 4 days the price of stock at ACME corporation fell $28. What was the average change in stock price per day?

Ludmilka [50]

The average change is 33 stock price per day

6 0

2 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

2 years ago

What’s the surface area

Nezavi [6.7K]

Answer:

Step-by-step explanation:

3 0

2 years ago

Some plese help i dont no what it is plese plese math

mr_godi [17]

Answer:

Factor: Something which is multiplied by something else. A factor can be a number, variable, term, or a longer expression. For example, the expression 7x(y+3) has three factors: 7 , x , and (y+3) .

3 0

2 years ago