Someone please help quick!

Measure the angle in degrees.

IrinaVladis [17]

Answer: 20 degrees

Have a good day :)

6 0

3 years ago

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

PLS HELP ME WILL GIVE BRAINLIEST

Gennadij [26K]

Answer:

31.2

Step-by-step explanation:

Hope it helps !!:)

3 0

3 years ago

(5/6)^-2 without exponent

slamgirl [31]

Answer:

36/25

Step-by-step explanation:

$(\frac{5}{6})^{-2}=(\frac{6}{5})^{2}\\(\frac{6}{5})^2=\frac{6}{5}\times\frac{6}{5}\\=\frac{36}{25}$

Steps:

Flip the fraction (negative exponent)

Square the fraction

7 0

3 years ago

Read 2 more answers

How do you solve these?

Murrr4er [49]

Answer:

Step-by-step explanation:

The first and only rule really is to factor these down to their primes and then apply a very simple rule

For every prime, take out 1 prime for every prime under the root sign that equals the index. The rest are thrown away.

That's very wordy. Let's try and see what it means with an example

Take sqrt(27) The index is 1/2 (square root) That means we need two threes in order to apply the rule.

sqrt(27) = sqrt(3 * 3 * 3 ) For every two primes take out 1 and throw one away.

sqrt(27) = 3 sqrt(3) You can't take out that 3rd 3.

64 = 2 * 2 *2 *2 *2 * 2

4th root 64 = 2*2*2 *2 * 2 *2

for every 4th root, you get to take 1 out and throw three away.

4th root 64 = 2 fourth root (2*2)

4th root 64 = 2 fourth root (4)

- 189 = - 3 * 3 * 3 * 7

cuberoot (- 189) = For every 3 roots, you get to pull 1 out and throw the other two away.

3 cube (- 7) is your answer.

72 = 2 * 2 * 2 * 3 * 3

cube root (72) = 2 cube root(9) You don't have enough threes to do any more than what is done.

6 0

3 years ago