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3 years ago

the area of a square game board is 144 square inches what is the length of one of the sides of the board

2 answers:

8 0

Each side is 36 inches. since it is a square all sides are the same length. 144 divided by 4 (number of sides) =36. hope that helps

slava [35]3 years ago

7 0

12 inches.

The other person who answered is wrong. He or she took it as perimeter.

Why 12 inches?

Because of
1) Square have equal sides
2) This is AREA not perimeter.

So to get 144 square inches, apply the formula for area of square...

Area of square = Length times Length

So 12 times 12, you will get 144.

Or simply use surd on calculator lol.

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why the first line starts from left to right on a cartesian plane, the second slants from top to bottom making it perpendicular

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2 years ago

Please helppp 3/4+(1/3÷1/6)-(-1/2)=

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2 years ago

Show work please<br> \sqrt(x+12)-\sqrt(2x+1)=1

Nesterboy [21]

Answer:

$x=4$

Step-by-step explanation:

Given $\displaystyle\\\sqrt{x+12}-\sqrt{2x+1}=1$ , start by squaring both sides to work towards isolating $x$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2$

Recall $(a-b)^2=a^2-2ab+b^2$ and $\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2\\\implies x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1$

Isolate the radical:

$\displaystyle\\x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1\\\implies -2\sqrt{(x+12)(2x+1)}=-3x-12\\\implies \sqrt{(x+12)(2x+1)}=\frac{-3x-12}{-2}$

Square both sides:

$\displaystyle\\(x+12)(2x+1)=\left(\frac{-3x-12}{-2}\right)^2$

Expand using FOIL and $(a+b)^2=a^2+2ab+b^2$ :

$\displaystyle\\2x^2+25x+12=\frac{9}{4}x^2+18x+36$

Move everything to one side to get a quadratic:

$\displaystyle-\frac{1}{4}x^2+7x-24=0$

Solving using the quadratic formula:

A quadratic in $ax^2+bx+c$ has real solutions $\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . In $\displaystyle-\frac{1}{4}x^2+7x-24$ , assign values:

$\displaystyle \\a=-\frac{1}{4}\\b=7\\c=-24$

Solving yields:

$\displaystyle\\x=\frac{-7\pm \sqrt{7^2-4\left(-\frac{1}{4}\right)\left(-24\right)}}{2\left(-\frac{1}{4}\right)}\\\\x=\frac{-7\pm \sqrt{25}}{-\frac{1}{2}}\\\\\begin{cases}x=\frac{-7+5}{-0.5}=\frac{-2}{-0.5}=\boxed{4}\\x=\frac{-7-5}{-0.5}=\frac{-12}{-0.5}=24 \:(\text{Extraneous})\end{cases}$

Only $x=4$ works when plugged in the original equation. Therefore, $x=24$ is extraneous and the only solution is $\boxed{x=4}$

4 0

2 years ago

Which part of the graph best represents the solution set to the system of inequalities y<_ x+1 and y+x<_ -1

Pavel [41]

Hello!

Remember that the symbols: ≤ and ≥ are graphed as a solid line. While the symbols: < and > are graphed as a dotted line.

Also, before graphing, it would be better to convert both equations to slope-intercept form.

y ≤ x + 1 is already in slope-intercept form.

y + x ≤ -1 is not written in slope-intercept form. (Slope-intercept form: y = mx + b)

y + x ≤ - 1 (subtract x from both sides)

y ≤ -x - 1

Graphing those lines, you get the graph below. You can see that Part C best represents the solution set systems of inequalities, because that is where both of the shaded lines intersect.

Answer: Part C

6 0

3 years ago

Read 2 more answers