The next larger whole number is 27 .
The next smaller whole number is 26 .

26.55 is closer to 27 than it is to 26 .

So the nearest whole number is 27 .

3 0

3 years ago

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There are 24 players on a team. Two of every six players were on the team last year. How

JulijaS [17]

8 players were on the team last year

3 0

2 years ago

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Suppose you are climbing a hill whose shape is given by the equation z = 1400 − 0.005x2 − 0.01y2, where x, y, and z are measured

Anarel [89]

Answer: You start to ascend

Explanation:

Since you are walking due south and the positive y-axis points north, the y-coordinate decreases.

Let

$y'$ = y-coordinate after walking due south
$z'$ = z-coordinate (or the height) after walking due south

So, the point where you stand after walking due south has a coordinate of $(100, y', z')$ . Moreover,

$z' = 1,400 - 0.05(100)^2 - 0.01(y')^2
\\ \boxed{z' = 1350 - 0.01(y')^2}$

Since the y-coordinate decreases after walking due south and $z' = 1350 - 0.01(y')^2$ ,

$y' \leq 120\\ \Rightarrow (y')^2 \leq 120^2 = 14,400\\ \Rightarrow -0.01(y')^2 \geq -0.01(14,400) = -144\\ \Rightarrow z' = 1,350 - 0.01(y')^2 \geq 1,350 - 0.01(14,400) = 1,206\\ \Rightarrow \boxed{z' \geq 1,206}$

So, the height after walking due south exceeds 1,206 meters, which is the previous height before walking due south. Therefore, if you walk due south from point with coordinates (100, 120, 1206), you are ascending.

5 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 $k$ :

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

$k \approx \pm 0.028$

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

7 0

2 years ago