What is the slope of the line that contains these points?

I don’t know how to find X or Y very urgent, need to figure this out how to solve plz help

Lena [83]

Answer:

x = 9

y = 9√3 = 15.6

Step-by-step explanation:

The triangle given is a right triangle, therefore:

✔️apply trigonometric ratio formula to find x:

Reference angle = 60°

Hypotenuse = 18

Adjacent = x

Thus:

Cos (60) = x/18

Multiply both sides by 18

18×cos(60) = x

9 = x

x = 9

✔️find y by applying pythagorean theorem:

y² = 18² - x² (pythagorean theorem)

y² = 18² - 9² (substitution)

y² = 243

y = √243

y = √(81*3)

y = 9√3 = 15.6

3 0

2 years ago

Answer quick and show work

shutvik [7]

Answer:

$15.19

Step-by-step explanation:

Write an equation

8.5% of 14 =

Convert percent into decimal:

8.5% = 0.085

"Of" means to multiply so multiply 0.085 to 14

0.085 x 14 = $1.19

14 + 1.19 = $15.19

8 0

3 years ago

Read 2 more answers

PLEASE HELP ME THANK YOU

Anna [14]

Answer:

m = -6/5

Step-by-step explanation:

The points present are (-2,2) and (3,-4).

m = -4 - 2 / 3 - (-2)

m = -4 - 2 / 3 + 2

m = -6/5

7 0

3 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

3 years ago

Help me solve this problem please

mel-nik [20]

It’s only a vertical angle

7 0

2 years ago