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3 years ago

I could really use a friend, it's been a rough day

Mathematics

2 answers:

AlladinOne [14]3 years ago

6 0

Answer:

i mean ill be ur friend if u want :)

Step-by-step explanation:

and i hope ur day gets better just know pain is temporary and swag is forever

Sunny_sXe [5.5K]3 years ago

4 0

I could be ur friend u can talk to me :)

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Plz help me with this like draw on a paper and tag a photo to this

ahrayia [7]

Answer:K to I Then K To K

Step-by-step explanation: x=7191028

3 0

3 years ago

Could someone help me?

Lady_Fox [76]

Step-by-step explanation:

The intercepted arc is twice the measure of the inscribed angle so if the inscribed angle BEC is 50°, then the measure of the arc BC is 100°.

8 0

3 years ago

A homogeneous rectangular lamina has constant area density ρ. Find the moment of inertia of the lamina about one corner

frozen [14]

Answer:

$I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

Step-by-step explanation:

By applying the concept of calculus;

the moment of inertia of the lamina about one corner $I_{corner}$ is:

$I_{corner} = \int\limits \int\limits_R (x^2+y^2) \rho d A \\ \\ I_{corner} = \int\limits^a_0\int\limits^b_0 \rho(x^2+y^2) dy dx$

where :

(a and b are the length and the breath of the rectangle respectively )

$I_{corner} = \rho \int\limits^a_0 {x^2y}+ \frac{y^3}{3} |^ {^ b}_{_0} \, dx$

$I_{corner} = \rho \int\limits^a_0 (bx^2 + \frac{b^3}{3})dx$

$I_{corner} = \rho [\frac{bx^3}{3}+ \frac{b^3x}{3}]^ {^ a} _{_0}$

$I_{corner} = \rho [\frac{a^3b}{3}+ \frac{ab^3}{3}]$

$I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

Thus; the moment of inertia of the lamina about one corner is $I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

7 0

3 years ago

Help me please!!! :((

snow_tiger [21]

Answer:

mark brainliest im grinding ranks

Step-by-step explanation:

4 0

3 years ago

What's the answer to this

stealth61 [152]

You need to replace x with 5 so it will be…
[3(5) + 4] - (2 - 5)
(15 + 4) - (-3)
19 + 3
22

3 0

4 years ago

Read 2 more answers