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2 years ago

I really need the help with this multiple choice question for geometry.

Mathematics

1 answer:

prisoha [69]2 years ago

4 0

Answer:

A. 50°

Step-by-step explanation:

The external angle ACB created by tangents CA and CB is the supplement of arc AB it intercepts.

∠ACB = 180° -AB

∠ACB = 180° -130°

∠ACB = 50°

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Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :

$\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}$

$k \approx \pm 0.028$

The positive value of $k$ will result in exactly one real root is approximately 0.028.

7 0

3 years ago

The graphs show the amounts in two different bank accounts over several months.

Neko [114]

Answer:

where are the graphs?

Step-by-step explanation:

3 0

3 years ago

Find the equation of the parabola that passes through

valentinak56 [21]

so we have the points of (0,-7),(7,-14),(-3,-19), let's plug those in the y = ax² + bx + c form, since we have three points, we'll plug each one once, thus a system of three variables, and then we'll solve it by substitution.

$\bf \begin{array}{cccllll} \stackrel{\textit{point (0,-7)}}{-7=a(0)^2+b(0)+c}& \stackrel{point (7,-14)}{-14=a(7)^2+b(7)+c}& \stackrel{point (-3,-19)}{-19=a(-3)^2+b(-3)+c}\\\\ -7=c&-14=49a+7b+c&-19=9a-3b+c \end{array}$

well, from the 1st equation, we know what "c" is already, so let's just plug that in the 2nd equation and solve for "b".

$\bf -14=49a+7b-7\implies -7=49a+7b\implies -7-49a=7b \\\\\\ \cfrac{-7-49a}{7}=b\implies \cfrac{-7}{7}-\cfrac{49a}{7}=b\implies -1-7a=b$

well, now let's plug that "b" into our 3rd equation and solve for "a".

$\bf -19=9a-3b-7\implies -12=9a-3b\implies -12=9a-3(-1-7a) \\\\\\ -12=9a+3+21a\implies -15=9a+21a\implies -15=30a \\\\\\ -\cfrac{15}{30}=a\implies \blacktriangleright -\cfrac{1}{2}=a \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{and since we know that}}{-1-7a=b}\implies -1-7\left( -\cfrac{1}{2} \right)=b\implies -1+\cfrac{7}{2}=b\implies \blacktriangleright \cfrac{5}{2}=b \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill y=-\cfrac{1}{2}x^2+\cfrac{5}{2}x-7~\hfill$