Answers? I have no clue what to do

Mathematics

2 answers:

liq [111]3 years ago

6 0

I would help but I took this class a while ago and don’t remember much... try the app MathPapa

zysi [14]3 years ago

3 0

#2 is answer 4, #3 is answer 3, #4 is answer 4. The answer to #5 is 6xy if it is a possibility.

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The diameter of the moon is 2,160 miles. A model has a scale of 1 in : 150 mi. What is the diameter of the model?

Charra [1.4K]

Answer:

The diameter of the model is 14.4 inches.

Step-by-step explanation:

The Diameter of the moon = 2,160 miles

The scale on the model represents 1 in = 150 miles

Let the model represents k inches in 2,160 miles.

So, by the Ratio of Proportionality:

$\frac{1}{150} = \frac{k}{2160}$

⇒ $k = \frac{2160}{150} \\\implies k = 14.4$

or, k = 14.4 inches

⇒On the scale 2160 miles is represented as 14.4 inches

Hence the diameter of the model is 14.4 inches.

3 0

3 years ago

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 $k$ :