Ask question

Ask question

All categories

3 years ago

6

9) Given an 8 foot length of tree trunk with a radius of 3 feet, what is the volume of the tree trunk section? (to the nearest w

hole ft3)

1 answer:

emmasim [6.3K]3 years ago

7 0

<span>226 is the answer.

</span>π×32<span>×8 = 226 </span>

You might be interested in

The sum of the square of a positive number and the square of 4 more than the number is 40. What is the number?

KengaRu [80]

Answer:

Step-by-step explanation:

2

6 0

2 years ago

F(x) = x + 2 g(x) = x – 4

wariber [46]

Answer:

G= -2/x, x $\neq$ 0

X= -2/g, g $\neq$ 0

Step-by-step explanation:

5 0

2 years ago

Find the missing side. Round to the nearest tenth.

Firdavs [7]

Answer: <em>23.7</em>

Step-by-step explanation:

<em>Let's calculate the acute angle at the base of the triangle:</em>

<em>180° - 90° - 40° = 50°</em>

<em>Find side x:</em>

$\dfrac{sin \: 90^{o} }{31} =\dfrac{sin \: 50^{o} }{x}\\\\sin \: 90^{o} \cdot x=sin \: 50^{o} \cdot 31\\\\sin \: 90^{o} =1\\sin \: 50^{o} \approx 0.766\\1 \cdot x =0.766 \cdot 31\\x=23.746\\23.746 \approx 23.7$

5 0

2 years ago

Andrew went to the store with $200.00. He bought 2 pair of shorts for $25.00 each, he bought 3 shirts for $22.00 each, and he bo

kap26 [50]

200-(25•2)=150
150-(22•3)=84
84-25=59
Andrew has $59 left

5 0

2 years ago

Read 2 more answers

Solve step by step solution then only i can do it plxx

Anuta_ua [19.1K]

Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°

So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle

So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

3 0

3 years ago