What should you do before buying a stock

The main show tank has a radius of 70 feet and forms a quarter sphere what is the volume of the quarter sphere shaped tank?

Usimov [2.4K]

Answer:

$359333.33\textrm{ }ft^{3}$

Step-by-step explanation:

Given:

Radius of spherical tank, $r=70\textrm{ ft}$

Volume of a quarter of a sphere is given as:

$V=\frac{1}{3}\pi r^{3}$

Plug in 70 ft for $r$ , $\frac{22}{7}$ for $\pi$ and solve for the volume, $V$ . This gives,

$V=\frac{1}{3}\pi r^{3}\\\\ V=\frac{1}{3}\times \frac{22}{7}\times (70)^{3}\\\\ V=359333.33\textrm{ }ft^{3}$

Therefore, the volume of the quarter sphere shaped tank is $359333.33\textrm{ }ft^{3}$

4 0

4 years ago

The arrow on the spinner is to be spun once.

Alex

The answer is 2/5. (Do not brainliest)

8 0

4 years ago

Read 2 more answers

WHAT is the scale factor of the dilation? PLEASE HELPPp c:

Artist 52 [7]

Answer:

2.5

Step-by-step explanation:

We can see that the original length of RS is 2 units. The length of R'S' is 5 units.

2 * x = 5

2x = 5

x = 5/2

x = 2.5

3 0

3 years ago

A magician’s stage has a rectangle trapdoor which is (x+ 1/2) by 2x ft. The outside rectangle length is (x+ 16) ft.

Katen [24]

Answer:

Part A:- $x +11 ft$ Part B:- $(4x+54)ft$ part C:- $x= 2ft$

Part D:- Yes it satisfy his requirement

Step-by-step explanation:

Given that,

length and breadth of Trapdoor are $(x+ \frac{1}{2} ) by 2x$ ft.

The Outside length of Rectangle is $(x+16 )$ ft.

Part A:- Total area (in square feet) of the stage can be represented by

$x^{2} + 27x+ 176$ . Write an expression for the width of stage.

So, Area of rectangle = $length \times breadth$

$x^{2} + 27x+ 176 = (x+16) \times breadth$

$Breadth = \frac{x^{2} + 27x+ 176}{x+16}$

$Breadth = x+11$ ft. ................(1)

Part B:- Write an expression for the Perimeter of the stage.

Here, Perimeter of Rectangle = $2(L+B)$

= $2(x+16 + x +11)$

= $(4x + 54) ft$

Part C:- The area of trapdoor is $10 ft^{2}$ . Find the value of x.

So, Area of trapdoor = $10 ft^{2}$

$(x +\frac{1}{2}) \times 2x =10 ft^{2}$

$2x^{2} +x - 10 = 0$

$x = \frac{-1-9}{4}$ , $\frac{-1+9}{4}$

$x = \frac{-5}{2}$ , $2$

Hence value of x is $2 ft$ (Neglecting the negative value because

length cannot be negative).

Part D:- The magician wishes to have the area of the stage be at least 20 times the area of the trapdoor. Does this stage satisfy his requirement? Explain.

So, Length of trapdoor is $(2+\frac{1}{2} ) = \frac{5}{2} ft$

breadth of trapdoor is $2\times 2 = 4ft$

Now, outside length of Rectangle is = $18ft$

And outside breadth of rectangle is = $13ft$

Here, Area of trapdoor = $\frac{5}{2} \times 4 = 10ft^{2}$ ...................(2)

Area of rectangle = $18 \times 13= 234 ft^{2}$ ....................(3)

Thus comparing Equation (1) & (2) we found that

Area of rectangular stage is 20 times greater than the area of trapdoor.

7 0

4 years ago

Use the graph of △ABC with midsegments DE, EF and DF. Show that EF is parallel to AC and that EF=1/2 AC

Vinvika [58]

According to the midsegment theorem, the midsegments are parallel to

and half the length of the opposite side.

The completed statement are as follows;

Because the slopes of $\overline{EF}$ and $\overline{AC}$ are both -4, $\overline{EF}$ ║ $\overline{AC}$

EF = $\underline{\sqrt{17}}$ and AC = $\underline{2 \cdot \sqrt{17} }$

Because $\underline{\sqrt{17} }$ = $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17} }$ , $EF = \frac{1}{2} \cdot AC$

Reasons:

From the given graph of ΔABC, we have;

Coordinates of the points A, B, and C are; A(-5, 2), B(1, -2), and C(-3, -6)

The coordinates of the point D and E on $\mathbf{\overline{DE}}$ are; D(-4, -2), and E(-2, 0)

The coordinates of the point F is; F(-1, -4)

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\displaystyle Slope \ of \ line \ \overline{AC} = \mathbf{\frac{(-6) - 2}{-3 - (-5)} = \frac{-8}{2}} = -4$

$\displaystyle Slope \ of \ line \ \overline{EF} = \frac{0 - (-4)}{-2 - (-1)} = \frac{4}{-1} = -4$

$Length \ of \ segment,\ l = \sqrt{\left (x_{2}-x_{1} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}$

Length of EF = √((-1 - (-2))² + (-4 - 0)²) = √(17)

Length of AC = √((-3 - (-5))² + (-6 - 2)²) = √(4 × 17) = 2·√(17)

Therefore, we have;

Because the slope of $\mathbf{\overline {EF}}$ and $\mathbf{\overline {AC}}$ are both , -4, $\overline {EF}$ ║ $\overline {AC}$ . EF = $\underline{\sqrt{17}}$ , and AC

= $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}$ ,. Because $\underline{\sqrt{17}}$ = $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}$ , $EF =\mathbf{ \frac{1}{2} AC}$

Learn more about midsegment theorem of a triangle here:

brainly.com/question/7423948

8 0

3 years ago