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2 years ago

2x+3y=12 find four solutions

Mathematics

1 answer:

Nana76 [90]2 years ago

7 0

Answer:

y = 0 and x = 6

y = 2 and x = 3

y = 4 and x = 0

y = 6 and x = -3

Step-by-step explanation:

2x+3y=12 is the equation of a straight line.

There are infinite number of solutions, but here you're probably looking for solutions where x and y are whole numbers. Trial and error will find you some, although if you examine the equation closely you see that if x is a multiple of 3 and y is a multiple of 2, then they produce terms that are the LCM (least common multiple) of 2 and 3, namely 6. That's the logic in the whole number solutions.

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A.75 cms B.64 cms C.90 cms D.125 cms

nirvana33 [79]

Answer:

Ans. 64 cm

Step-by-step explanation:

we have to use the ratio of actual distance and distance on projection here.

Given,

Ratio of distance in projection and actual distance is 1:200,000

That means, 1 cm in projection will be 200,000 cm in actual distance.

Here,

Actual distance is 128 km.

we know, 1 km=100,000 cm

Now, 128 km=128,00,000 cm

So, actual distance is 128 00 000 cm.

Since, 200000 cm in actual distance means 1 cm in projection.

So,

for 12800000 cm in actual distance means 12800000/200000 cm in projection.

So, in the screen distance will be-

12800000 / 200000

=64 cm

3 0

3 years ago

Read 2 more answers

Find $\int \dfrac{x}{\sqrt{1-x^4}}$ Please, help

ki77a [65]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2867785

_______________

Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-x^4}}\,dx}\\\\\\ \mathsf{=\displaystyle\int\! \frac{1}{2}\cdot 2\cdot \frac{1}{\sqrt{1-(x^2)^2}}\,dx}\\\\\\ \mathsf{=\displaystyle \frac{1}{2}\int\! \frac{1}{\sqrt{1-(x^2)^2}}\cdot 2x\,dx\qquad\quad(i)}$

Make a trigonometric substitution:

$\begin{array}{lcl}
\mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\
&&\mathsf{t=arcsin(x^2)\,,\qquad 0\ \textless \ x\ \textless \ \frac{\pi}{2}}\end{array}$

so the integral (i) becomes

$\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{1-sin^2\,t}}\cdot cos\,t\,dt\qquad\quad (but~1-sin^2\,t=cos^2\,t)}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{cos^2\,t}}\cdot cos\,t\,dt}$

$\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{cos\,t}\cdot cos\,t\,dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\,t+C}$

Now, substitute back for t = arcsin(x²), and you finally get the result:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=\frac{1}{2}\,arcsin(x^2)+C}$ ✔

________

You could also make

x² = cos t

and you would get this expression for the integral:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=-\,\frac{1}{2}\,arccos(x^2)+C_2}$ ✔

which is fine, because those two functions have the same derivative, as the difference between them is a constant:

$\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\
=\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}$

$\mathsf{=\dfrac{\pi}{4}}$ ✔

and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.

I hope this helps. =)

6 0

2 years ago

Find the surface area of the triangular prism (above) using its net (below).

r-ruslan [8.4K]

Answer:

96 square units

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one. Please have a look at the attached photo.

My answer: