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ruslelena [56]
2 years ago
14

PLEASE HELP! ILL GIVE BRAINIEST!!!

Mathematics
1 answer:
jasenka [17]2 years ago
8 0
The answer would be

Fraction 3/10
Percent 30%
Decimal 0.3

There is 10 parts and it ask for probability of 3 numbers which would be 3/10
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Find the surface area of the net below.
Hatshy [7]

Answer:

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5 0
2 years ago
What is the scientific notation for 10.6x10^-11?
Lina20 [59]
000000.0000000000106
3 0
3 years ago
Question 1 of 10
melisa1 [442]

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4 0
2 years ago
Can someone thoroughly explain this implicit differentiation with a trig function. No matter how many times I try to solve this,
Anton [14]

Answer:

\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}

Step-by-step explanation:

So we have the equation:

\tan(x-y)=\frac{y}{8+x^2}

And we want to find dy/dx.

So, let's take the derivative of both sides:

\frac{d}{dx}[\tan(x-y)]=\frac{d}{dx}[\frac{y}{8+x^2}]

Let's do each side individually.

Left Side:

We have:

\frac{d}{dx}[\tan(x-y)]

We can use the chain rule, where:

(u(v(x))'=u'(v(x))\cdot v'(x)

Let u(x) be tan(x). Then v(x) is (x-y). Remember that d/dx(tan(x)) is sec²(x). So:

=\sec^2(x-y)\cdot (\frac{d}{dx}[x-y])

Differentiate x like normally. Implicitly differentiate for y. This yields:

=\sec^2(x-y)(1-y')

Distribute:

=\sec^2(x-y)-y'\sec^2(x-y)

And that is our left side.

Right Side:

We have:

\frac{d}{dx}[\frac{y}{8+x^2}]

We can use the quotient rule, where:

\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}

f is y. g is (8+x²). So:

=\frac{\frac{d}{dx}[y](8+x^2)-(y)\frac{d}{dx}(8+x^2)}{(8+x^2)^2}

Differentiate:

=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}

And that is our right side.

So, our entire equation is:

\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}

To find dy/dx, we have to solve for y'. Let's multiply both sides by the denominator on the right. So:

((8+x^2)^2)\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}((8+x^2)^2)

The right side cancels. Let's distribute the left:

\sec^2(x-y)(8+x^2)^2-y'\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy

Now, let's move all the y'-terms to one side. Add our second term from our left equation to the right. So:

\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy+y'\sec^2(x-y)(8+x^2)^2

Move -2xy to the left. So:

\sec^2(x-y)(8+x^2)^2+2xy=y'(8+x^2)+y'\sec^2(x-y)(8+x^2)^2

Factor out a y' from the right:

\sec^2(x-y)(8+x^2)^2+2xy=y'((8+x^2)+\sec^2(x-y)(8+x^2)^2)

Divide. Therefore, dy/dx is:

\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)+\sec^2(x-y)(8+x^2)^2}

We can factor out a (8+x²) from the denominator. So:

\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}

And we're done!

8 0
3 years ago
Graph and write solution:<br><br> 2x + 1 = y<br><br> -4x + 6y = -10
PIT_PIT [208]

The graph is shown in figure below

The Solution Set is (-2,-3)

Step-by-step explanation:

We need to graph the equations and writ the solution.

For graphing we need to find the values of x and y.

For that, we need to solve the given equations:

2x + 1 = y\\-4x + 6y = -10

Let:

2x + 1 = y\,\,\,eq(1)\\-4x + 6y = -10\,\,\,eq(2)

We can solve this by using Substitution Method.

Putting value of y of eq(1) into eq(2) and finding value of x:

2x + 1 = y\,\,\,eq(1)\\-4x+6(2x+1)=-10\\-4x+12x+6=-10\\8x+6=-10\\8x=-10-6\\8x=-16\\x=-16/8\\x=-2

So, value of x = -2

Now put value of x in eq(1) to find value of y:

2x + 1 = y\\2(-2)+1=y\\-4+1=y\\y=-3

So, value of y = -3

Plotting on graph: x=-2 and y = -3

The graph is shown in figure below

The Solution Set is (-2,-3)

Keywords: graph the equations

Learn more about Graphing Equations at:

  • brainly.com/question/10541435
  • brainly.com/question/3126500
  • brainly.com/question/2334270

#learnwithBrainly

4 0
3 years ago
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