Find the perfect square trinomial x^2+26x+c

1 answer:

4 0

Use the formula (b/2)^2 in order to create a new term in order to find the perfect square trinomial. Your answer should 169 or I’m completely wrong just saying

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Of the students in your class, 12% are left-handed. What fraction of the students are left-handed? Are there more right-handed o

kvv77 [185]

The left-handed students
12% are left handed, percent means per hundred. Change percent into per 100.
12% = $\dfrac{12}{100}$

Then simplify the fraction by dividing the numerator and denominator by 4
$\dfrac{12}{100}$
= $\dfrac{12:4}{100:4}$
= $\dfrac{3}{25}$
This is the fraction showing the proportion of the left-handed students

The right-handed students
Determine the fraction showing the proportion of the right-handed students
1 - the left-handed
= 1 - $\dfrac{3}{25}$

Equalize the denominator to 25
$\dfrac{25}{25} - \dfrac{3}{25}$

Simplify
$\dfrac{21}{25}$
This is the proportion of the right-handed students

<span>Are there more right-handed or left-handed students?
There are more right-handed students than the left-handed students, because 21/25 is more than 3/25</span>

4 0

3 years ago

Kat is painting the edge of a triangular stage prop with reflective orange paint. The lengths of the edges of the triangle are (

Eva8 [605]

Plug 4 in for every x

(3*4)-4=8
(4^2)-1=15 [(4^2) is 4*4]
(2*(4^2))-15= 15 [because of PEMDAS the exponent comes first and then multiply by 2 then subtract]

Perimeter is the sum of the lengths so add up all the sides from above
8+15+15=38

3 0

3 years ago

The area of the triangle formed by x− and y− intercepts of the parabola y=0.5(x−3)(x+k) is equal to 1.5 square units. Find all p

Juliette [100K]

Check the picture below.

based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).

and that will give us two x-intercepts, at x = 3 and x = k.

since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.

now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.

let's find the y-intercept by setting x = 0 now,

$\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)$

$\bf \cfrac{3}{2}=\cfrac{3+k}{2}\left( -\cfrac{3k}{2} \right)\implies \stackrel{\textit{multiplying by }\stackrel{LCD}{2}}{3=\cfrac{(3+k)(-3k)}{2}}\implies 6=-9k-3k^2 \\\\\\ 6=-3(3k+k^2)\implies \cfrac{6}{-3}=3k+k^2\implies -2=3k+k^2 \\\\\\ 0=k^2+3k+2\implies 0=(k+2)(k+1)\implies k= \begin{cases} -2\\ -1 \end{cases}$

now, we can plug those values on A = (1/2)bh,

$\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5$