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Vilka [71]
3 years ago
8

A road sign has the shape of an equilateral triangle. If the sides of the sign measure 25 in., what is the height of the sign to

the nearest tenth of an inch? please help out please
Mathematics
1 answer:
Sever21 [200]3 years ago
4 0

Answer: socratic app

Step-by-step explanation:

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If anyone can help me I’ll give all my points to u..
balu736 [363]

Answer: 12x squared

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
PLEASE ANSWER THIS
kumpel [21]
(30 0Z less then) is your answer :)
6 0
3 years ago
Read 2 more answers
Simplify, state all restrictions.
Kipish [7]

The simplified expression is \frac{1 - x - y}{x + y}and the restriction is y \ne -x

<h3>How to simplify the expression?</h3>

The expression is given as:

\frac{x - y}{4x^2 - 8xy + 3y^2} \div \frac{2x + y}{2x - 3y} \times \frac{4x^2 - y^2}{x^2 - y^2} -1

Express x^2 - y^2 as (x + y)(x - y) and factorize other expressions

\frac{x - y}{(2x - y)(2x - 3y)} \div \frac{2x + y}{2x - 3y} \times \frac{4x^2 - y^2}{(x - y)(x + y)} -1

Rewrite the expression as products

\frac{x - y}{(2x - y)(2x - 3y)} \times \frac{2x - 3y}{2x + y} \times \frac{4x^2 - y^2}{(x - y)(x + y)} -1

Cancel out the common factors

\frac{1}{(2x - y)} \times \frac{1}{2x + y} \times \frac{4x^2 - y^2}{(x + y)} -1

Express 4x^2 - y^2 as (2x - y)(2x + y)

\frac{1}{(2x - y)} \times \frac{1}{2x + y} \times \frac{(2x - y)(2x + y)}{(x + y)} -1

Cancel out the common factors

\frac{1}{x + y} -1

Take the LCM

\frac{1 - x - y}{x + y}

Hence, the simplified expression is \frac{1 - x - y}{x + y}and the restriction is y \ne -x

Read more about expressions at:

brainly.com/question/723406

#SPJ1

7 0
1 year ago
Can someone please help me with this question?!? I am so confused and I don't know how to answer it.
lbvjy [14]

9514 1404 393

Answer:

  a. f(0) = 1

  b. DNE (does not exist)

  c. DNE

  d. lim = 3

Step-by-step explanation:

The function exists at a point if it is defined there. The function is defined anywhere on the solid line and at solid dots. It is not defined at open circles. So, the function is defined everywhere except (2, 3), which has an open circle.

The open circle at (0, 4) prevents the function from being doubly-defined at x=0, since it is already defined to be 1 at x=0.

This discussion tells you ...

  f(0) = 1

 f(2) does not exist. There is a "hole" in the function definition there.

__

The function has a limit at a point if approaching from the left and approaching from the right have you approaching that same point.

Consider the point (1, 2). The graph is a solid line through that point. Approaching from values less than x=1, we get to the same point (1, 2) as when we approach from values greater than x=1.

Similarly, consider the point (2, 3). Approaching from values of x less than 2, we get to the same point (2, 3) as when we approach from x-values greater than 2. The limit at x=2 is 3. The only difference from the previous case is that the function is not actually defined to be that value there.

__

Now consider what happens at x=0. When we approach from the left, we approach the point (0, 4). When we approach from the right, we approach the point (0, 1). These are different points. Because they are different coming from the left and from the right, we say "the limit as x→0 does not exist."

__

In summary, ...

  a) f(0) = 1

  b) lim x → 0 does not exist

  c) f(2) does not exist

  d) lim x → 2 = 3

_____

<em>Additional comment</em>

The significance of the function not being defined at a point where the limit exists, (2, 3), is that <em>the function is not continuous there</em>. This kind of discontinuity is called "removable", because we could make the function continuous at x=2 by defining f(2) = 3 (that is, "filling the hole").

6 0
2 years ago
A circle has a radius of 14 units. What is the area in square units of the circle in terms of π?
Snowcat [4.5K]
Area of circle = pi r²

14²= 196

196 x pi = 196pi

Area = 196pi units²
3 0
3 years ago
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