All categories

2 years ago

Please help!

Mathematics

2 answers:

coldgirl [10]2 years ago

7 0

Answer:

Question 1 )

given ,

height of cylinder = 9 cm

diameter of cylinder = 10 cm

$\dashrightarrow \: radius = \frac{10}{5} \: cm \\$

now ,

$Volume \: of \: cylinder = \pi \: r {}^{2} h \\ \\ substitutng \: the \: values \: of \: h \: and \: r \: \\ \\ \dashrightarrow \: \pi \: \times (\frac{10}{2} ) {}^{2} \times 9 \\ \\ \dashrightarrow \: \pi \: \times 25 \times 9 \\ \\ \dashrightarrow \: 225\pi \: cm {}^{3}$

therefore ,

option (C) is correct.

______________________________________

Question - 2 )

dimensions of the rectangular prism are 2 × 3 × 4

$Lateral \: surface \: area = 2h(l + b) \\ \\ \dashrightarrow \: 2 \times 3(4 + 2) \\ \\ \dashrightarrow \: 6(6) \\ \\ \dashrightarrow \: 36 \: cm {}^{2}$

hope helpful :D

Iteru [2.4K]2 years ago

6 0

The volume of the cylinder is 225 $\rm cm^3$ and the lateral surface area of the rectangular prism is 36 square cm.

It is given that the radius of the cylinder is 10 cm and height is 9 cm.

It is required to find the volume of the cylinder.

<h3>What is volume?</h3>

It is defined as a three-dimensional space enclosed by an object or thing.

We know the volume of a cylinder given by:

$\rm V = \pi r^2 h$

We have diameter = 10 cm

Radius r = 10/2 = 5 cm

and height h = 9 cm

Volume will be:

$\rm V= \pi \times5^2 \times 9$

V = 225π cubic cm

For the lateral surface area = 2h(l+b)

= 2×3(4+2)

= 36 square cm

Thus, the volume of the cylinder is 225 $\rm cm^3$ and the lateral surface area of the rectangular prism is 36 square cm.

Learn more about the volume here:

brainly.com/question/16788902

You might be interested in

18.50x2.6 what is the answer helppppp!!!!!!!!!!!!!!

ser-zykov [4K]

Answer:

48.1

Explanation:

Remove Decimals & Multiply

1850 · 26

= 4810

18.5 Has One Decimal Places

2.6 Has One Decimal Place

Move The Decimal Over Two Places

4810 $\rightarrow$ 48.1

6 0

3 years ago

Read 2 more answers

12,11,28,27,44,43,60,59 what number pattern?

Alinara [238K]

05 22 –27 08 40 37 –03 22 –22 –23 –02 –20 48 53 18 31 –41 –38 –29 –47 30 – 13 46 –07 29 –06 –40 – 19 –20 02 22 20 –06 17 35 31 –01 –37 –34 – 18 –30 ... 27 28 10 —ll 16 10 –50 –52 – 10 — 13 35 40 27 28 –54 –70 –36 –39 –48 –05 03 03 – 12 –24 – 13 – 14 – 14 –46 –29 –40 39 60 38 48 –59 –59–43 –27 –44 –01 ...ik this didnt help but ill never know if it acually did

6 0

3 years ago

Read 2 more answers

Turn the sentence into an expression 5.three times the sum of two consecutive integers

ryzh [129]

3(x+x+1)

hope this helps

8 0

3 years ago

Which table shows a proportional relationship?

storchak [24]

Answer:

the answer is d because the numbers are multiples of eachother

Step-by-step explanation:

3 0

3 years ago

(1 point) (a) Find the point Q that is a distance 0.1 from the point P=(6,6) in the direction of v=⟨−1,1⟩. Give five decimal pla

natima [27]

Answer:

following are the solution to the given points:

Step-by-step explanation:

In point a:

$\vec{v} = -\vec{1 i} +\vec{1j}\\\\|\vec{v}| = \sqrt{-1^2+1^2}$

$=\sqrt{1+1}\\\\=\sqrt{2}$

calculating unit vector:

$\frac{\vec{v}}{|\vec{v}|} = \frac{-1i+1j}{\sqrt{2}}$

the point Q is at a distance h from P(6,6) Here, h=0.1

$a=-6+O.1 \times \frac{-1}{\sqrt{2}}\\\\= 5.92928 \\\\b= 6+O.1 \times \frac{-1}{\sqrt{2}} \\\\= 6.07071$

the value of Q= (5.92928 ,6.07071 )

In point b:

Calculating the directional derivative of $f (x, y) = \sqrt{x+3y}$ at P in the direction of $\vec{v}$

$f_{PQ} (P) =\fracx{f(Q)-f(P)}{h}\\\\$

$=\frac{f(5.92928 ,6.07071)-f(6,6)}{0.1}\\\\=\frac{\sqrt{(5.92928+ 3 \times 6.07071)}-\sqrt{(6+ 3\times 6)}}{0.1}\\\\= \frac{0.197651557}{0.1}\\\\= 1.97651557$

$\vec{v} = 1.97651557$

In point C:

Computing the directional derivative using the partial derivatives of f.

$f_x(x,y)= \frac{1}{2 \sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{2 \sqrt{22}}\\\\f_x(x,y)= \frac{1}{\sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{\sqrt{22}}\\\\f_{(PQ)}(P)= (f_x \vec{i} + f_y \vec{j}) \cdot \frac{\vec{v}}{|\vec{v}|}\\\\= (\frac{1}{2 \sqrt{22}}\vec{i} + \frac{1}{\sqrt{22}} \vec{j}) \cdot \frac{-1}{\sqrt{2}}\vec{i} + \frac{1}{\sqrt{2}} \vec{j}$

4 0

3 years ago