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3 years ago

Given: m =-2/3

Mathematics

1 answer:

Elenna [48]3 years ago

6 0

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The correct answer is (( D )) .

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A right triangle has leg lengths of 9 units and 12 units. Write an equation that can be

LenKa [72]

<h2>The Pythagorean Theorem</h2>

The Pythagorean Theorem: $a^2+b^2=c^2$

<em>a</em> and <em>b</em> are the two legs in a right triangle
<em>c</em> is the hypotenuse (the longest side of a right triangle; opposite the 90° angle
<em>*keep in mind that this theorem applies only to right triangles</em>

<h2>Solving the Question</h2>

We're given:

Legs = 9 units and 12 units
Hypotenuse = <em>c</em>

<em />

Plug these measurements into the Pythagorean theorem:

$a^2+b^2=c^2\\9^2+12^2=c^2$

⇒ Combine like terms:

$81+144=c^2\\c^2=225$

<h2>Answer</h2>

$c^2=225$

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

Robin can mow a lawn in 3 hours, while Brady can mow the same lawn in 4 hours. How many hours would it take for them to mow the

BaLLatris [955]

The answer is <span>twelve sevenths (12/7)
</span>
x - the job<span>
t - time to do the job x

</span><span>Robin can mow a lawn in 3 hours: x/3 * t
</span><span>Brady can mow the same lawn in 4 hours: x/4 * t

</span><span>If they work together, the total work will be the sum of their speeds:
</span>x = x/3 * t + x/4 * t = (x/3 + x/4)*t = (4x/12 + 3x/12) * t = 7x/12 * t

x = 7x/12 * t

Divide both sides by x:
x/x = 7x/12 * t/x
1 = 7/12 t
t = 1 : 7/12
t = 12/7

5 0

3 years ago

An experiment consists of rolling a six-sided die to select a number between 1 and 6 and drawing a card at random from a set of

Readme [11.4K]

<span>The event definition that corresponds to exactly one outcome of the experiment is that both numbers are 5s.</span>

5 0

2 years ago

Read 2 more answers

PLEASE HELP WILL MARK U BRAINLIEST

larisa86 [58]

Answer: x=27

Step-by-step explanation:

3 0

3 years ago