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2 years ago

What are the measures of angles M and N?

Mathematics

1 answer:

romanna [79]2 years ago

7 0

Answer:

mM = 113 and mN = 61

Step-by-step explanation:

In a Cyclic quadrilateral, the rule states that:

The sum opposite interior angles is equal to 180°

In the above diagram, we have cyclic quadrilateral KLMN

According to the rule stated above:

Angle K is Opposite to Angle M

So, Angle K + Angle M = 180°

Angle K = 67°

67° + Angle M = 180°

Angle M = 180° - 67°

Angle M = 113°

Angle L is Opposite to Angle N

so Angle L + Angle N = 180°

Angle L is given as = 119°

119° + Angle N = 180°

Angle N = 180° - 119°

Angle N = 61°

Therefore, Angle M = 113° and Angle N = 61°

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Read 2 more answers

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.

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