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RUDIKE [14]
2 years ago
15

Lily buys 378 crystals for 6 dollars, What is the cost of 109 crystals?

Mathematics
1 answer:
miskamm [114]2 years ago
3 0
About $1.73. Since 378 cost six dollars, The price of one equates to be about 0.0158. Using this knowledge, we can multiply by 109 to get roughly 1.7301
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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

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3 years ago
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Answer:

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hope this helps....

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