Define a variable and write an expression for the phrase.

Jkl is an equilateral triangle jk= 13x+5,KL= 17X-19 AND JL= 8X+35

stira [4]

Answer:

x = 6

Step-by-step explanation:

Since the triangle is equilateral then sides are congruent.

Equate any 2 sides and solve for x

13x + 5 = 8x + 35 ( subtract 8x from both sides )

5x + 5 = 35 ( subtract 5 from both sides )

5x = 30 ( divide both sides by 5 )

x = 6

JK = (13 × 6) + 5 = 78 + 5 = 83

KL = (17 × 6) - 19 = 102 - 19 = 83

JL = (8 × 6) + 35 = 48 + 35 = 83

ΔJKL is equilateral with side = 83

4 0

2 years ago

Definition: The ___ of an event happening is expressed as a fraction between 0 and 1.

aliina [53]

Definition: The <u>probability</u> of an event happening is expressed as a fraction between 0 and 1.

Example: The <u>chance</u> that you roll a 4 on a die is 1

3 0

3 years ago

A ball is thrown straight out at 80 feet per second from an upstairs window that's 15 feet off the ground. Find the ball's horiz

Alex73 [517]

Answer:

Step-by-step explanation:

In order to find the horizontal distance the ball travels, we need to know first how long it took to hit the ground. We will find that time in the y-dimension, and then use that time in the x-dimension, which is the dimension in question when we talk about horizontal distance. Here's what we know in the y-dimension:

a = -32 ft/s/s

v₀ = 0 (since the ball is being thrown straight out the window, the angle is 0 degrees, which translates to no upwards velocity at all)

Δx = -15 feet (negative because the ball lands 15 feet below the point from which it drops)

t = ?? sec.

The equation we will use is the one for displacement:

Δx = $v_0t+\frac{1}{2}at^2$ and filling in:

$-15=(0)t+\frac{1}{2}(-32)t^2$ which simplifies down to

$-15=-16t^2$ so

$t=\sqrt{\frac{-15}{-16} }$ so

t = .968 sec (That is not the correct number of sig fig's but if I use the correct number, the answer doesn't come out to be one of the choices given. So I deviate from the rules a bit here out of necessity.)

Now we use that time in the x-dimension. Here's what we know in that dimension specifically:

a = 0 (acceleration in this dimension is always 0)

v₀ = 80 ft/sec

t = .968 sec

Δx = ?? feet

We use the equation for displacement again, and filling in what we know in this dimension:

Δx = $(80)(.968) +(0)(.968)^2$ and of course the portion of that after the plus sign goes to 0, leaving us with simply: