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gladu [14]
3 years ago
6

I know the answer to this question but how many solutions does it have?

Mathematics
2 answers:
BlackZzzverrR [31]3 years ago
8 0

Answer:Hence, x = 5, y = 0 is the required solution.

Step-by-step explanation:

Masteriza [31]3 years ago
6 0

Answer:

There is only one solution

Step-by-step explanation:

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The equation y=1/2x represents a proportional relationship. What is the constant of proportionality ?
Helen [10]

Answer:

Is that the full question?

Step-by-step explanation:

6 0
3 years ago
Y=square root of 6x-5/4x please answer before Wednesday
Wewaii [24]

Answer:

x = 5/6-4y^2

Step-by-step explanation:

^ = to the power of and * = times

y = square root of 6x-5/4x

(times by the square root on either side, sorry I don't know how to word it)

y^2=6x-5/4x

(times by 4x either side)

4xy^2= 6x-5

(+5 either side)

4xy^2+5=6x

(-4xy^2 either side)

5=6x-4xy^2

(factorize)

5=x(6-4y^2)

(divide by (6-4y^2))

5/6-4y^2 = x     or     x = 5/6-4y^2

Tell me if this is incorrect.

5 0
3 years ago
Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4
arsen [322]

we are given

R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k

now, we can find x , y and z components

x=cos(8t),y=sin(8t),z=8ln(cos(t))

Arc length calculation:

we can use formula

L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt

x'=-8sin(8t),y=8cos(8t),z=-8tan(t)

now, we can plug these values

L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt

now, we can simplify it

L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt

L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt

L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt

L=\int _0^{\frac{\pi }{4}}8sec(t) dt

now, we can solve integral

\int \:8\sec \left(t\right)dt

=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|

now, we can plug bounds

and we get

=8\ln \left(\sqrt{2}+1\right)-0

so,

L=8\ln \left(1+\sqrt{2}\right)..............Answer

5 0
3 years ago
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