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Bess [88]
3 years ago
15

Kyla makes a triangular school pennant. The area of the triangle is 180 square inches. The base of the pennant is z inches long.

The height is 6 inches longer than twice the base length. What is the hieght of the pendant if a=1/2(b)(h)
Mathematics
1 answer:
kykrilka [37]3 years ago
4 0
We can represent the base as z and the height as 2z+6. We are going to use the formula A=1/2*b*h and solve for z
180=1/2*z*(2z+6)
360=2z^2+6z 
0=2z^2+6z-360
0=2(z^2+3z-180)
0=(z+15)(z-12)
So z=-15 and 12 but it must be positive so then the base is equal to 12

When we plug this into 2z+6 we get 30 for the height
2(12)+6=30

Hope this helps
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Find the vertical and horizontal asymptotes, domain, range, and roots of f (x) = -1 / x-3 +2.
gladu [14]

Answer:

Vertical asymptote: x=3

Horizontal asymptote: f(x) =2

Domain of f(x) is all real numbers except 3.

Range of f(x) is all real numbers except 2.

Step-by-step explanation:

Given:

Function:

f (x) = -\dfrac{1 }{ x-3} +2

One root, x = 3.5

To find:

Vertical and horizontal asymptote, domain, range and roots of f(x).

Solution:

First of all, let us find the roots of f(x).

<em>Roots of f(x) means the value of x where f(x) = 0</em>

0= -\dfrac{1 }{ x-3} +2\\\Rightarrow 2= \dfrac{1 }{ x-3}\\\Rightarrow 2x-2 \times 3=1\\\Rightarrow 2x=7\\\Rightarrow x = 3.5

One root, x = 3.5

Domain of f(x) i.e. the values that we give as input to the function and there is a value of f(x) defined for it.

For x = 3, the value of f(x) \rightarrow \infty

For all, other values of x , f(x) is defined.

Hence, Domain of f(x) is all real numbers except 3.

Range of f(x) i.e. the values that are possible output of the function.

f(x) = 2 is not possible in this case because something is subtracted from 2. That something is \frac{1}{x-3}.

Hence, Range of f(x) is all real numbers except 2.

Vertical Asymptote is the value of x, where value of f(x) \rightarrow \infty.

-\dfrac{1 }{ x-3} +2 \rightarrow \infty

It is possible only when

x-3=0\\\Rightarrow x=3

\therefore vertical asymptote: x=3

Horizontal Asymptote is the value of f(x) , where value of x \rightarrow \infty.

x\rightarrow \infty \Rightarrow \dfrac{1 }{ x-3} \rightarrow 0\\\therefore f(x) =-0+2 \\\Rightarrow f(x) =2

\therefore Horizontal asymptote: f(x) =2

Please refer to the graph of given function as shown in the attached image.

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