

To solve these type of problems you need to use the pythagoras theorem ⇨ Hypotenuse² = Base² + Altitude².
Here,
- Altitude = 1.6 cm.
- Base = 1.2 cm
- Hypotenuse = x
Now, let's solve for x.
Hypotenuse² = Base² + Altitude²
x² = (1.2)² + (1.6)²
x² = 1.44 + 2.56
x² = 4
x = √4
x = <em><u>2</u></em><em><u>.</u></em>
- So, the value of x is <em><u>2</u><u> </u><u>cm.</u></em>
<h3>
<u>NOTE</u><u> </u><u>:</u><u>-</u></h3>
- Pythagoras theorem can be used only in the cases of right-angled triangles. Here, it's given that the triangle is right angled so we can use this theorem.
- To solve the squares if decimals, take them as whole numbers & then just add the decimal points. For example, ⇨ for (1.2)², take it as 12² , then multiply 12 by 12, you'll get 144. Now, add the decimal place accordingly ⇨ 1.44 . So, (1.2)² = 1.44.
Answer:
<h3>Graph 3</h3>
Line starting at x = -2
- <u>Domain</u>: x ≥ -2
- <u>Range</u>: y ≥ 0
<h3>Graph 4</h3>
Vertical line
- <u>Domain</u>: x = 3
- <u>Range</u>: y = any real number
<h3>Graph 5</h3>
Quadratic function with negative leading coefficient and max value of 3
- <u>Domain</u>: x = any real number
- <u>Range</u>: y ≤ 3
<h3>Graph 6</h3>
Curve with non-negative domain and min value of -2
- <u>Domain</u>: x ≥ 0
- <u>Range</u>: y ≥ -2
<h3>Graph 7</h3>
Line with no restriction
- <u>Domain</u>: x = any real number
- <u>Range</u>: y = any real number
<h3>Graph 8</h3>
Quadratic function with positive leading coefficient and min value of 4
- <u>Domain</u>: x = any real number
- <u>Range</u>: y ≥ 4
<h3>Graph 9</h3>
Parabola with restriction at x = -4
- <u>Domain</u>: x = any real number except -4
- <u>Range</u>: y = any real number
<h3>Graph 10</h3>
Square root function with star point (2, 0)
- <u>Domain</u>: x ≥ 2
- <u>Range</u>: y ≥ 0
A unit rate where one is in the Denometer, so you would want to divide the bottom by 10.
Multiply -12 to each of the terms (make sure to carry the negative!)
-36a - 24b + 12 will be your final answer :)