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3 years ago

Find the perimeter of the triangle

Mathematics

1 answer:

ArbitrLikvidat [17]3 years ago

4 0

Answer:

60 ft

Step-by-step explanation:

(x+13) = (3x-1) = (6x-22) since the triangle is equaililateral

(x+13) = (3x-1) (we only need to chose 2)

-2x=-14

x=7

after this, we just need to plug in the number for x

3(7)-1= 20--->since the triangle has 3 congruent sides, they all have the distance, so 20x3=60

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Please help!

BaLLatris [955]

a) The polynomial in expanded form is $f(x) = x^{3}-2\cdot x^{2}-21\cdot x -18$ .

b) The slant asymptote is represented by the linear function is $y = -x + 1$ .

c) There is a discontinuity at $x = 2$ with a slant asymptote.

a) In this question we are going to use the Factor Theorem, which establishes that polynomial are the result of products of binomials of the form $x-r_{i}$ , where $r_{i}$ is the i-th root of the polynomial and the grade is equal to the quantity of roots. Therefore, the polynomial $f(x)$ has the following form:

$f(x) = (x-6)\cdot (x+1)\cdot (x+3)$

And the expanded form is obtained by some algebraic handling:

$f(x) = (x-6)\cdot (x^{2}+4\cdot x +3)$

$f(x) = x\cdot (x^{2}+4\cdot x + 3)-6\cdot (x^{2}+4\cdot x +3)$

$f(x) = x^{3}+4\cdot x^{2}+3\cdot x -6\cdot x^{2}-24\cdot x -18$

$f(x) = x^{3}-2\cdot x^{2}-21\cdot x -18$ (1)

The polynomial in expanded form is $f(x) = x^{3}-2\cdot x^{2}-21\cdot x -18$ .

b) In this question we divide the polynomial found in a) (in factor form) by the polynomial $x^{2}-x -2$ (also in factor form). That is:

$g(x) = \frac{(x-6)\cdot (x+1)\cdot (x+3)}{(x-2)\cdot (x+1)}$

$g(x) = \frac{(x-6)\cdot (x+3)}{x-2}$ (2)

The slant asymptote is defined by linear function, whose slope ( $m$ ) and intercept ( $b$ ) are determined by the following expressions:

$m = \lim_{x \to \pm \infty} \frac{g(x)}{x}$ (3)

$b = \lim_{x \to \pm \infty} [g(x)-x]$ (4)

If $g(x) = \frac{(x-6)\cdot (x+3)}{x-2}$ , then the equation of the slant asymptote is:

$m = \lim_{x \to 2} \frac{(x-6)\cdot (x+3)}{x\cdot (x-2)}$

$m = \lim_{x \to \pm \infty} \left(\frac{x^{2}-3\cdot x -18}{x^{2}-2\cdot x} \right)$

$m = 1$

$b = \lim_{x \to \pm \infty} \left(\frac{x^{2}-3\cdot x -18}{x-2}-x \right)$

$b = \lim_{x \to \pm \infty} \left(\frac{x^{2}-3\cdot x - 18-x^{2}+2\cdot x}{x-2}\right)$

$b = \lim_{n \to \infty} \left(\frac{-x-18}{x-2} \right)$

$b = -1$

The slant asymptote is represented by the linear function is $y = -x + 1$ .

c) The number of discontinuities in rational functions is equal to the number of binomials in the denominator, which was determined in b). Hence, we have a discontinuity at $x = 2$ with a slant asymptote.

We kindly invite to check this question on asymptotes: brainly.com/question/4084552

8 0

2 years ago

Which boxes have a volume of 120 cubic feet? (4 points)

sergiy2304 [10]

Answer:

A,D

Step-by-step explanation:

5x4x6=120

10x3x4=120

5 0

3 years ago

Read 2 more answers

√48a^4 b^5 c^3 d simplify and keep it in radical form

Morgarella [4.7K]

I will assume the square root extends all the way across.

$\sqrt{48a^4b^5c^3}$

$\sqrt{2^4\cdot3\cdot a^4b^5c^3}$

$\sqrt{(2^2)^2\cdot3\cdot(a^2)^2\cdot(b^2)^2\cdot b\cdot c^2\cdot c}$

$\sqrt{(4)^2\cdot3\cdot(a^2)^2\cdot(b^2)^2\cdot b\cdot c^2\cdot c}$

$\sqrt{(4\cdot a^2 \cdot b^2 \cdot c)^2\cdot3\cdot b\cdot c}$

$4\cdot a^2\cdot b^2\cdot c\sqrt{3\cdot b\cdot c}$

$\boxed{4a^2b^2c\sqrt{3bc}}$

3 0

4 years ago

What’s the answer to this question??

katen-ka-za [31]

I’d love to help I pretty sure it is C-6

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3 years ago

Which rational expression has a value of 0 when x = 2 ? A) 7x -5/x2 + 10 B) -3x +6/8x-9 C) -5x + 8 / x-2

shepuryov [24]

answer for this question is (b) - 3 x + 6 x /8x- 9

7 0

3 years ago