Write the equation for the following quadratic relation in standard form:

Find the measure of the supplement of the angle. 62° Enter your answer in the box

iren2701 [21]

Supplementary angles =180

180-62=118

6 0

3 years ago

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9.0 3.5 140 ----- 12 12 units?

xz_007 [3.2K]

Mdndkddjntrhd hccjdjdj the dodo

7 0

3 years ago

Charles asked his teammates how many hours they practiced swimming during the week. He recorded his data in the table below and

Kitty [74]

Answer:

The answer is C.

Step-by-step explanation:

7 0

3 years ago

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The values in the table represent a function.

sweet [91]

Answer:

f(-6) = 8

f(3) = -2

f(x) = -5 when x is 4

Step-by-step explanation:

* Lets explain how to solve the problem

- The table of the function has two column

# First column labeled x with entries:

-6 , 7 , 4 , 3 , -5

# Second column labeled f(x) with entries:

8 , 3 , -5 , -2 , 12

∴ The ordered pairs of the function f(x) are:

(-6 , 8) , (7 , 3) , (4 , -5) , (3 , -2) , (-5 , 12)

* Lets complete the missing

∵ The value of x in the first row is -6

∵ The value of f(x) in the first row is 8

∴ The function notation in the 1st row is f(-6) = 8

- The ordered pair given in the first row of the table can be

written using function notation as f(-6) = 8

∵ The ordered pair whose x = 3 is (3 , -2)

∴ The value of f(x) when x = 3 is -2

∴ f(3) = -2

∵ The ordered pair whose f(x) = -5 is (4 , -5)

∴ The value of x when f(x) = -5 is 4

∴ f(x) = -5 when x is 4

6 0

3 years ago

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Find the point P on the graph of the function y=√x closest to the point (9,0)

Sphinxa [80]

Answer:

$\displaystyle \frac{17}{2}$ .

Step-by-step explanation:

Let the $x$ -coordinate of $P$ be $t$ . For $P\!$ to be on the graph of the function $y = \sqrt{x}$ , the $y$ -coordinate of $\! P$ would need to be $\sqrt{t}$ . Therefore, the coordinate of $P \!$ would be $\left(t,\, \sqrt{t}\right)$ .

The Euclidean Distance between $\left(t,\, \sqrt{t}\right)$ and $(9,\, 0)$ is:

$\begin{aligned} & d\left(\left(t,\, \sqrt{t}\right),\, (9,\, 0)\right) \\ &= \sqrt{(t - 9)^2 +\left(\sqrt{t}\right)^{2}} \\ &= \sqrt{t^2 - 18\, t + 81 + t} \\ &= \sqrt{t^2 - 17 \, t + 81}\end{aligned}$ .

The goal is to find the a $t$ that minimizes this distance. However, $\sqrt{t^2 - 17 \, t + 81}$ is non-negative for all real $t\!$ . Hence, the $\! t$ that minimizes the square of this expression, $\left(t^2 - 17 \, t + 81\right)$ , would also minimize $\sqrt{t^2 - 17 \, t + 81}\!$ .

Differentiate $\left(t^2 - 17 \, t + 81\right)$ with respect to $t$ :

$\displaystyle \frac{d}{dt}\left[t^2 - 17 \, t + 81\right] = 2\, t - 17$ .

$\displaystyle \frac{d^{2}}{dt^{2}}\left[t^2 - 17 \, t + 81\right] = 2$ .

Set the first derivative, $(2\, t - 17)$ , to $0$ and solve for $t$ :

$2\, t - 17 = 0$ .

$\displaystyle t = \frac{17}{2}$ .

Notice that the second derivative is greater than $0$ for this $t$ . Hence, $\displaystyle t = \frac{17}{2}$ would indeed minimize $\left(t^2 - 17 \, t + 81\right)$ . This $t\!$ value would also minimize $\sqrt{t^2 - 17 \, t + 81}\!$ , the distance between $P$ $\left(t,\, \sqrt{t}\right)$ and $(9,\, 0)$ .

Therefore, the point $P$ would be closest to $(9,\, 0)$ when the $x$ -coordinate of $P\!$ is $\displaystyle \frac{17}{2}$ .

8 0

3 years ago