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3 years ago

6

Please help! what is the measure of x?

1 answer:

Alexeev081 [22]3 years ago

4 0

28°
Since KL and FG are perpendicular they create a 90° angle. 180-90-62=28

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A square field has an area of 121 square metre what is the length of each side

Tresset [83]

30.25 square meters because you would divide 121 by 4 because a square has 4 equal length sides.
Hope this helps

3 0

3 years ago

IgorC [24]

Answer:

12 less than or equal to 2

Step-by-step explanation:

7 0

3 years ago

To two decimal places, find the value of k that will make the function f(x) continuous everywhere. f of x equals the quantity 3x

Fiesta28 [93]

Given function is

$f(x)=\left\{\begin{matrix}3x+k & x\leq -4 \\ kx^2-5 & x> -4\end{matrix}\right.$

now we need to find the value of k such that function f(x) continuous everywhere.

We know that any function f(x) is continuous at point x=a if left hand limit and right hand limits at the point x=a are equal.

So we just need to find both left and right hand limits then set equal to each other to find the value of k

To find the left hand limit (LHD) we plug x=-4 into 3x+k

so LHD= 3(-4)+k

To find the Right hand limit (RHD) we plug x=-4 into

$kx^2-5$

so RHD= $k(-4)^2-5$

Now set both equal

$k(-4)^2-5=3(-4)+k$

$16k-5=-12+k$

$16k-k=-12+5$

$15k=-7$

$k=-\frac{7}{15}$

k=-0.47

<u>Hence final answer is -0.47.</u>

7 0

3 years ago

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4

dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

$x_1\ =\ -2$

$y_1\ =\ 3$

$z_1\ =\ -4$

Distance is measure across the line

$\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}$

So, we can write

$\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k$

$=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k$

$=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k$

$=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}$

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

$3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3$

$=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18$

$=>18k-24+12k+9+10k-32\ =\ 18$

$=>\ k\ =\dfrac{13}{8}$

So,

$x\ =\ 3k-4$

$=\ 3\times \dfrac{13}{8}-4$

$=\ \dfrac{7}{4}$

$y\ =\ \dfrac{4k+3}{2}$

$=\ \dfrac{4\times \dfrac{13}{8}+3}{2}$

$=\ \dfrac{19}{4}$

$z\ =\ \dfrac{5k-16}{3}$

$=\ \dfrac{5\times \dfrac{13}{8}-16}{3}$

$=\ \dfrac{-21}{8}$

Now, the distance of point from the plane is given by,

$d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$

$=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}$

$=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}$

$=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}$

$=\ \sqrt{\dfrac{1177}{64}}$

$=\ 4.28$

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

= 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0

3 years ago

PLEASE HELP ME NOW URGENT

timama [110]

ANSWER

$y = 3 \pm \sqrt{21}$

EXPLANATION

The quadratic equation is:

${y}^{2} - 6y - 12 = 0$

Group variable terms:

${y}^{2} - 6y = 12$

Add the square of half, the coefficient of y to both sides.

${y}^{2} - 6y + ( - 3) ^{2} = 12 + ( - 3) ^{2}$

${y}^{2} - 6y + 9= 12 + 9$

The LHS us now a perfect square trinomial:

${(y - 3)}^{2}= 21$

Take square root:

$y - 3 = \pm \sqrt{21}$

$y = 3 \pm \sqrt{21}$

The first choice is correct.

8 0

3 years ago

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