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3 years ago

Solve each system by graphing.

Mathematics

1 answer:

Kryger [21]3 years ago

3 0

9514 1404 393

Answer:

(x, y) = (4, -4)

Step-by-step explanation:

A graphing calculator makes graphing very easy. The attachment shows the solution to be (x, y) = (4, -4).

The equations are in slope-intercept form, so it is convenient to start from the y-intercept and use the slope (rise/run) to find additional points on the line.

The first line can be drawn by staring at (0, -2) and moving down 1 grid unit for each 2 to the right.

The second line can be drawn by starting at (0, 2) and moving down 3 grid units for each 2 to the right.

The point of intersection of the lines, (4, -4), is the solution to the system of equations.

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Let x the age of John
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x+(x+9)= 23
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3 years ago

Given:

diamong [38]

Answer:

10-5 $\sqrt{2}$

Step-by-step explanation:

As per the attached figure, right angled $\triangle MDL$ has an inscribed circle whose center is $I$ .

We have joined the incenter $I$ to the vertices of the $\triangle MDL$ .

Sides MD and DL are equal because we are given that $\angle M = \angle L = 45 ^\circ.$

Formula for area of a $\triangle = \dfrac{1}{2} \times base \times height$

As per the figure attached, we are given that side a = 10.

Using pythagoras theorem, we can easily calculate that side ML = 10 $\sqrt{2}$

Points P,Q and R are at $90 ^\circ$ on the sides ML, MD and DL respectively so IQ, IR and IP are heights of $\triangle$ MIL, $\triangle$ MID and $\triangle$ DIL.

Also,

$\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL$

$\dfrac{1}{2} \times 10 \times 10 = \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10\sqrt2\\\Rightarrow r = \dfrac {10}{2+\sqrt2} \\\Rightarrow r = \dfrac{5\sqrt2}{\sqrt2+1}\\\text{Multiplying and divinding by }(\sqrt2 +1)\\\Rightarrow r = 10-5\sqrt2$

So, radius of circle = $10-5\sqrt2$

8 0

3 years ago

￼ Solve each system by graphing.

Solve each system by graphing.