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SOVA2 [1]
3 years ago
13

Really need helpppp?!

Mathematics
1 answer:
user100 [1]3 years ago
5 0

Answer:

6.43

pls follow me and mark brainliest ...thank u

Step-by-step explanation:

length of side x will be

Tan tita = opp/adjacent

tan65 = x/3

tan 65 is 2.14

so...you are left with 2.14=x/3

by cross multiplying

x= 2.14 × 3

= 6.43

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We have a right triangle that has am 8 meter and 21 meter leg .Find the hypotenuse
Serjik [45]

Answer:

hypotenuse is 22.47 m

Step-by-step explanation:

The length of both legs of a right angle triangle are 8m and 21 m

We need to find the hypotenuse

To find hypotenuse we use Pythagorean theorem

Hypotenuse is AC  and other two legs are AB  and BC

AC^2 = AB^2 + BC^2

Hypotenuse ^2 = 8^2 + 21^2

hypotenuse = \sqrt{8^2 + 21^2}

= \sqrt{64 + 441}

= \sqrt{505}

= 22.47

So the length of the hypotenuse is 22.47 m

4 0
3 years ago
Given the Homogeneous equation x^2ydy+xy^2dx=0, use y=ux, u=y/x and dy=udx+xdu to solve the differential equation. Solve for y.
scZoUnD [109]

Answer:

y=\frac{C}{x}.

Step-by-step explanation:

Given homogeneous equation

x^2ydy+xy^2dx=0

\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{xy^2}{x^2y}

Substitute y=ux , u=\frac{y}{x}

\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{y}{x}

Now,

u+x\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}x}

u+x\frac{\mathrm{d}u}{\mathrm{d}x}=-u

\frac{\mathrm{d}u}{\mathrm{d}x}=-2u

\frac{du}{u}=-\frac{dx}{x}

Integrating both side we get

lnu=-2lnx+lnC

Where lnC= integration constant

lnu+ln{x}^2=lnC

lnux^2=lnC

Cancel ln on both side

ux^2=C

Substitute u=\frac{y}{x}

Then we get

xy=C

y=\frac{C}{x}.

Answer:y=\frac{C}{x}.

8 0
3 years ago
3. A square metal plate has a density of 10.2 g/cm3 and weighs 2.193 kg.
MakcuM [25]

Answer:

215 cm³

8.6 cm

Step-by-step explanation:

given,

density = 10.2 g/cm³

mass = 2.193 kg = 2193g

a. Volume = mass / density

=2193 / 10.2

= 215 cm³

b. given that base area = 25cm²

volume = base area x thickness

thickness = volume / base area

= 215 / 25

= 8.6 cm

4 0
3 years ago
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3 years ago
I need help with Math please
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