Vertical asymptote and horizontal asymptote

I need help please tell me the answer you will get 30 points too. I. 7x+42

noname [10]

What exactly are you looking for

7 0

3 years ago

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Pls respond fast I’m tired lol I can’t figure it out

AURORKA [14]

Answer:

62°

Step-by-step explanation:

∠IKD=∠EIB=∠GJI =118° (Corresponding angles)

∠GJI+∠AJG=180° (Linear Pair)

∴∠AJG= 180°-118°

=62°

6 0

3 years ago

Incline mats, or triangle mats, are offered with different levels of incline to help gymnasts learn basic moves. As the name may

nata0808 [166]

Answer:

The length of the mat is 30 inches.

Step-by-step explanation:

Let H be the hypotenuse of the incline mat, L its length and h its height.

It is given that H = L + 4 and h = L - 14. Now, using Pythagoras' theorem for right-angled triangles, H² = L² + h²

(L + 4)² = L² + (L - 14)²

Expanding, we have

L² + 8L + 16 = L² + L² -28L + 196

collecting all the terms to the right side, we have

L² + L² -28L + 196 - L² - 8L - 16 = 0

L² - 36L + 180 = 0

Using the quadratic formula, L =

L $\frac{-(-36) +/- \sqrt{(-36)^{2} -4X1X180} )}{2X1} = \frac{36 +/- \sqrt{1296 - 720} }{2} \\= \frac{36 +/- \sqrt{576} }{2} \\= \frac{36 +/- 24 )}{2}\\=\frac{36 + 24}{2} or \frac{36 - 24}{2} \\=\frac{60}{2} or \frac{12}{2} \\= 30 or 6$

We use the larger answer L = 30 inches since L = 6 would give h = L -14 = 6 - 14 = -8 .

So, the length of the mat is 30 inches.

3 0

3 years ago

Passing through (2, - 2) and perpendicular to the line whose equation is y= 5x+2

jonny [76]

Step-by-step explanation:

Hey there!!!

Here,

Given, A line passes through point (2,-2) and is perpendicular to the y= 5x+2.

The equation of a straight line passing through point is,

$(y - y1) = m1(x - x1)$

Now, put all values.

$(y + 2) = m1(x - 2)$

It is the 1st equation.

Another equation is;

y = 5x +2........(2nd equation).

Now, Comparing it with y = mx + c, we get;

m2=5

As per the condition of perpendicular lines,

m1×m2= -1

m1 × 5 = -1

Therefore, m2= -1/5.

Keeping the value of m1 in 1st equation.

$(y + 2) = \frac{ - 1}{5} (x - 2)$

Simplify them.

$5(y + 2) = - x + 2$

$5y + 10 = - x + 2$

$x + 5y + 8 = 0$

Therefore the required equation is x+5y+8= 0.

Hope it helps...

7 0

4 years ago

What is the area of a regular octagon with a side length of 5 m and a distance from the center to the vertex of 6.5 m ?

horsena [70]

Answer is 120
Use the Pythagorean theorem to find the distance from the center to the side, 6.5²-(5/2)²=6², so the distance is 6. The perimeter is 5×8=40 then use ½(6)(40)=120

7 0

4 years ago

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