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What is the approximate volume of a basketball with the circumference of 42ft ?

Semmy [17]

We know that

[volume of a sphere]=(4/3)*pi*r³

length of a circumference=42 ft
[length of a circumference]=2*pi*r-------> r=[length of a circumference]/(2*pi)
r=[42]/(2*pi)---------> r=6.6845 ft

[volume of a sphere]=(4/3)*pi*6.6845³-----------> 1251.11 ft³

the answer is 1251.11 ft³

4 0

2 years ago

Name the triangles that are classified by angles

Bumek [7]

<span>2. scalene, isosceles, equilateral</span>

3 0

2 years ago

Read 2 more answers

Help I’ll give you 30 points

Gnesinka [82]

Answer:

a) 16.25 centimeters

b) 6.8 centimeters

Step-by-step explanation:

We can set up a proportion for both of these problems.

a) Look at the longest side of each triangle. We can set up a fraction: 7.2/18. We can also set up a fraction for XZ and PR:

6.5/x

x represents the length you're trying to find.

Since the triangles are similar, we have an equation:

7.2/18=6.5/x

Cross multiply

16.25

XZ is 16.25 centimeters long.

b)

We can do same thing:

7.2/18=x/17

Cross multiply

6.8

QR is 6.8 centimeters long.

Hope this helps!

6 0

2 years ago

Read 2 more answers

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Sliva [168]

Answer:

Step-by-step explanation:

5 0

2 years ago

The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg

julia-pushkina [17]

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

Let $\rm O$ denote the observer.
Let $\rm A$ denote the top of the tree.
Let $\rm R$ denote the base of the tree.
Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$ .

Angles:

Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$ .
Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$ .

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$ .

The question is asking for the length of segment $\rm AB$ . Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$ .

The length of segment $\rm OB$ could be represented in two ways:

In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$ .
In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$ .

For example, in right triangle $\rm \triangle OBR$ , the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$ . The length of that segment is $x\; \rm ft$ .

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$ .

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$ ) in terms of $x$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$ .

Similarly, in right triangle $\rm \triangle OBA$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$ .

Equate the right-hand side of these two equations:

$0.810\, x \approx 0.649\, (x + 20)$ .

Solve for $x$ :

$x \approx 81\; \rm ft$ .

Hence, the height of the top of this tree relative to the base of the hill would be $(x + 20)\; {\rm ft}\approx 101\; \rm ft$ .

6 0

2 years ago