We can see that triangle JKM is a right triangle, with right angle at M. Segment KM is 6 units and segment MJ is 3 units. We can also see that KJ is hypotenuse of right triangle.

We will use Pythagoras theorem to solve for KJ as:

$KJ^2=KM^2+MJ^2$

$KJ^2=6^2+3^2$

$KJ^2=36+9$

$KJ^2=45$

Now we will take positive square root on both sides:

$\sqrt{KJ^2}=\sqrt{45}$

$KJ=\sqrt{9\cdot 5}$

$KJ=3\sqrt{5}$

Therefore, the length of line segment KJ is $3\sqrt{5}$ and option D is the correct choice.

5 0

4 years ago

Find the area of the circle below???

olga55 [171]

Answer:

Area of the circle = $24.618\,feet^2$

Step-by-step explanation:

Radius of the circle= $2.8\,feet$

Area of the circle= $\pi \times r^2$

As,

$\pi =\dfrac{22}{7} =3.14$

Area of the circle is:

$=3.14\times 2.8 \times 2.8\\\\= 3.14 \times 7.84\\\\=24.618\,feet^2$

The Area of the circle is : $24.618\,feet^2$

5 0

3 years ago

Read 2 more answers

HELP URGENT PLEASE!!!!!!!!!!!!!!

leonid [27]

Answer:

The supplement of 89° is the angle that when added to 89° forms a straight angle (180° ).

Step-by-step explanation:

4 0

3 years ago

Select the two values of x that are roots of this equation.<br> 2x^2+ 1 = 5x

iren [92.7K]

Answer:

$\frac{5+\sqrt{17}}{4}$ , $\frac{5-\sqrt{17}}{4}$

Step-by-step explanation:

One is asked to find the root of the following equation:

$2x^2+1=5x$

Manipulate the equation such that it conforms to the standard form of a quadratic equation. The standard quadratic equation in the general format is as follows:

$ax^2+bx+c=0$

Change the given equation using inverse operations,

$2x^2+1=5x$

$2x^2-5x+1=0$

The quadratic formula is a method that can be used to find the roots of a quadratic equation. Graphically speaking, the roots of a quadratic equation are where the graph of the quadratic equation intersects the x-axis. The quadratic formula uses the coefficients of the terms in the quadratic equation to find the values at which the graph of the equation intersects the x-axis. The quadratic formula, in the general format, is as follows:

$\frac{-b(+-)\sqrt{b^2-4ac}}{2a}$

Please note that the terms used in the general equation of the quadratic formula correspond to the coefficients of the terms in the general format of the quadratic equation. Substitute the coefficients of the terms in the given problem into the quadratic formula,

$\frac{-b(+-)\sqrt{b^2-4ac}}{2a}$

$\frac{-(-5)(+-)\sqrt{(-5)^2-4(2)(1)}}{2(2)}$

Simplify,

$\frac{-(-5)(+-)\sqrt{(-5)^2-4(2)(1)}}{2(2)}$

$\frac{5(+-)\sqrt{25-8}}{4}$

$\frac{5(+-)\sqrt{17}}{4}$

Rewrite,

$\frac{5(+-)\sqrt{17}}{4}$

$\frac{5+\sqrt{17}}{4}$ , $\frac{5-\sqrt{17}}{4}$

8 0

3 years ago