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2 years ago

PLEASE HELP! ILL GIVE BRAINIEST!!!

1 answer:

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:

okay so first to find the surface area of the net below you should always start from the line that is pointing at the end of the the i n u m so you see that is between 8 1/8 and 2 so you see the line you draw a line in the middle between 8 and then when you look and then you draw another line were 19 years and then you draw another line so you can see that at the top of 19 and there at the bottom of 918 you see that you can find the Nets below because you can see me you draw your line it shows you Drawing the Line because when you draw a circle in the middle and then you draw a another little circle in the middle is shows the surface area of the net below so bye

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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]

Answer:

The answer is D AC because that line is opposite of angle B

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:

#learnwithBrainly

4 0

3 years ago