3x-y=-7 2x + 3y<br><img src="https://tex.z-dn.net/?f=3x%20-%20y%20%3D%20%20-%207%20%5C%20%5C%5C%202x%20%2B%203y%20%3D%2012" id="

Solve<br>3 tan 8 <-1 for 0 5 0521.

BlackZzzverrR [31]

Answer:

Substitute the given value into the function and evaluate.

3 tan ( 8 ) < − 1 = 0.4216225 < − 1

0 = 0.4216225 < − 1

5 = 0.4216225 < − 1

521 = 0.4216225 < − 1

Step-by-step explanation:

3 0

2 years ago

) Find the coordinates of the point A' which is the symmetric point

soldier1979 [14.2K]

Let, coordinate of point A' is (x,y).

Since, A' is the symmetric point A(3, 2) with respect to the line 2x + y - 12 = 0.

So, slope of line containing A and A' will be perpendicular to the line 2x + y - 12 = 0 and also their center lies in the line too.

Now, their center is given by :

$C( \dfrac{x+3}{2}, \dfrac{y+2}{2})$

Also, product of slope will be -1 .( Since, they are parallel )

$\dfrac{y-2}{x-3} \times -2 = -1\\\\2y - 4 = x - 3\\\\2y - x = 1$

x = 2y - 1

So, $C( \dfrac{2y -1 +3}{2}, \dfrac{y+2}{2})\\\\C( \dfrac{2y + 2}{2}, \dfrac{y+2}{2})$

Also, C satisfy given line :

$2\times ( \dfrac{2y + 2}{2}) + \dfrac{y+2}{2} = 12\\\\4y + 4 + y + 2 = 24\\\\5y = 18\\\\y = \dfrac{18}{5}$

Also,

$x = 2\times \dfrac{18}{5 } - 1\\\\x = \dfrac{31}{5}$

Therefore, the symmetric points is $A'(\dfrac{31}{5}, \dfrac{18}{5})$ .

7 0

2 years ago

In the word problem below, what answer or solution are you being asked to

sweet-ann [11.9K]

Answer:

the answer is C 6x2 .....................

8 0

2 years ago

mrs.beluga is driving on a snow covered road with a drag factor of 0.2. she brakes suddenly for a deer. The tires leave a yaw ma

Aneli [31]

First, we are going to find the radius of the yaw mark. To do that we are going to use the formula: $r= \frac{c^2}{8m} + \frac{m}{2}$
where
$c$ is the length of the chord
$m$ is the middle ordinate
We know from our problem that the tires leave a yaw mark with a 52 foot chord and a middle ornate of 6 feet, so $c=52$ and $m=6$ . Lets replace those values in our formula:
$r= \frac{52^2}{8(6)} + \frac{6}{2}$
$r= \frac{2704}{48} +3$
$r= \frac{169}{3} +3$
$r= \frac{178}{3}$

Next, to find the minimum speed, we are going to use the formula: $s= \sqrt{15fr}$
where
$f$ is <span>drag factor
</span> $r$ is the radius
We know form our problem that the drag factor is 0.2, so $f=0.2$ . We also know from our previous calculation that the radius is $\frac{178}{3}$ , so $r= \frac{178}{3}$ . Lets replace those values in our formula:
$s= \sqrt{(15)(0.2)( \frac{178}{3}) }$
$s= \sqrt{178}$
$s=13.34$ mph

We can conclude that Mrs. Beluga's minimum speed before she applied the brakes was 13.34 miles per hour.

8 0

3 years ago

I don’t understand what i can do

alexandr402 [8]

Answer:

14.037

Step-by-step explanation:

$3^{-3} +5+9\\$

0.037+5+9

14.037

4 0

2 years ago