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2 years ago

Please helppp i will mark brainlistt!!!!!!

Mathematics

1 answer:

Citrus2011 [14]2 years ago

6 0

Answer:

26 square meters ♡♡♡♡♡♡♡♡♤◇◇♤£¥

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Name the property illustrated by each statement:

aniked [119]

both are correct.

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8 0

2 years ago

Identify the correct corresponding parts

andrey2020 [161]

Answer:

The correct corresponding part is;

$\overline {CB}$ ≅ $\overline {CD}$

Step-by-step explanation:

The information given symbolically in the diagram are;

ΔCAB is congruent to ΔCED (ΔCAB ≅ ΔCED)

Segment $\overline {CA}$ is congruent to $\overline {CE}$ ( $\overline {CA}$ ≅ $\overline {CE}$ )

Segment $\overline {CB}$ is congruent to $\overline {CD}$ ( $\overline {CB}$ ≅ $\overline {CD}$ )

From which, we have;

∠A ≅ ∠E by Congruent Parts of Congruent Triangles are Congruent (CPCTC)

∠B ≅ ∠D by CPCTC

Segment $\overline {AB}$ is congruent to $\overline {DE}$ ( $\overline {AB}$ ≅ $\overline {DE}$ ) by CPCTC

Segment $\overline {AE}$ bisects $\overline {BD}$

Segment $\overline {BD}$ bisects $\overline {AE}$

Therefore, the correct option is $\overline {CB}$ ≅ $\overline {CD}$

3 0

2 years ago

Find the area a and b circumference of the circle with radius 10 m and use 3.14 for pi

zhenek [66]

Answer:

$\rule6cm99999cm$

6 0

2 years ago

Read 2 more answers

I don't know the answer can eney one help??

zlopas [31]

Answer:

Please see attached picture for full solution.

4 0

2 years ago

the volume v of a right circular cylinder of radius r and heigh h is V = pi r^2 h 1. how is dV/dt related to dr/dt if h is const

laiz [17]

In general, the volume

$V=\pi r^2h$

has total derivative

$\dfrac{\mathrm dV}{\mathrm dt}=\pi\left(2rh\dfrac{\mathrm dr}{\mathrm dt}+r^2\dfrac{\mathrm dh}{\mathrm dt}\right)$

If the cylinder's height is kept constant, then $\dfrac{\mathrm dh}{\mathrm dt}=0$ and we have

$\dfrac{\mathrm dV}{\mathrm dt}=2\pi rh\dfrac{\mathrm dt}{\mathrm dt}$

which is to say, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ are directly proportional by a factor equivalent to the lateral surface area of the cylinder ( $2\pi r h$ ).

Meanwhile, if the cylinder's radius is kept fixed, then

$\dfrac{\mathrm dV}{\mathrm dt}=\pi r^2\dfrac{\mathrm dh}{\mathrm dt}$

since $\dfrac{\mathrm dr}{\mathrm dt}=0$ . In other words, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dh}{\mathrm dt}$ are directly proportional by a factor of the surface area of the cylinder's circular face ( $\pi r^2$ ).

Finally, the general case ( $r$ and $h$ not constant), you can see from the total derivative that $\dfrac{\mathrm dV}{\mathrm dt}$ is affected by both $\dfrac{\mathrm dh}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ in combination.

8 0

2 years ago