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Dmitry_Shevchenko [17]

2 years ago

15

Triangle END is reflected across the line y = x using the rule (x, y) → (y, x) to create triangle E′N′D′. If a line segment is d

rawn from point E to point E′, which statement would best describe the line segment drawn in relation to the line y = x?

1 answer:

maw [93]2 years ago

7 0

Answer:

They are perpendicular.

Step-by-step explanation:

For example, if we have the point E (1, 2) and E' (2, 1)

we can get the slope as follows

m' = (yE' - yE) / (xE' - xE)

⇒ m' = (1 - 2)/(2 - 1) = -1 / 1 = -1

Now we have the slope of the line y = x, which is m = 1

we can apply

m*m' = (1)*(-1) = - 1

we can say that they are perpendicular.

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14. circunferencia = 7.3 m (calcular<br>el radio ) *<br>

Tanya [424]

Answer:

radius =1.1618m

Step-by-step explanation:

circumference of circle =7.3m

2πr=7.3

r=7.3/(2π)=1.1618m

6 0

2 years ago

Tia and her 5 friends are going to the ice skating ring. Each person pays $5 for admission and $5 for food. Write a numerical ex

dedylja [7]

The expression would be (5*6)+(5*6)=x
or you could shorten it and do (10*6)=x

8 0

2 years ago

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

2 years ago

Please answer this! I really suck at math and I need to do well.

Mariulka [41]

I believe it would be G.

5 0

2 years ago

. Here is an equation 2x + 9 = -15. Write three different equations that have the same solution as 2x + 9 = -15. Show or explain

Anettt [7]

Answer:

1. $x-3=-15$
2. $2x=-24$
3. $2x-3=-27$

Step-by-step explanation:

Given the following question:

$2x+9=-15$

To find three equations that have the same solution as the given equation we have to first solve the equation.

$2x+9=-15$
$9-9=0$
$-15-9=-24$
$2x=-24$
$2x\div2=x$
$-24\div2=-12$
$x=-12$

<u>Equation one:</u>
$x-3=-15$
$-3+3=0$
$-15+3=-12$
$x=-12$

<u>Equation two:
</u> $2x=-24$
$2x\div2=x$
$-24\div2=-12$
$x=-12$

<u>Equation three:</u>
$2x-3=-27$
$-3+3=0$
$-27+3=-24$
$-24\div2=-12$
$x=-12$

Hope this helps.

4 0

1 year ago