Answer:
- Question 1: x=55
- Question 2: m∠A=45°
Step-by-step explanation:
<u>Question #1</u>
- Since it is an isosceles triangle, this means that the base angles have the same value.
- The angle sum is 180°
Solve
2x+35+35=180
2x+70=180
2x=110
x=55
<u>Question #2</u>
- Since it is given there is a right angle, this means the angle is 90°
- The angle sum is 180°
Solve
5x+5x+90=180
10x+90=180
10x=90
x=9
It asked to find m∠A, which is 5x. So 5x=5(9)=45°
Hope this helps!! :)
Please let me know if you have any questions
Answer:
Answer <u>B</u>
Step-by-step explanation:
The formula for the area off a triangle is 1/2Bh
The base of the triangle is Z, and the height is W, so the formula is
A
When doing translations the object is not changing size only the position in which they are in. Circle P is still the same size as Circle R so no dilation was used, only transformation
Basically you can graph a function, for example a parabola by following the step pattern 1,35...
If you take the "standard" parabola, y = x², which has it's vertex at the origin (0, 0), then:
<span>➊ one way you can use a "step pattern" is as follows: </span>
<span>Starting from the vertex as "the first point" ... </span>
<span>OVER 1 (right or left) from the vertex point, UP 1² = 1 from the vertex point </span>
<span>OVER 2 (right or left) from the vertex point, UP 2² = 4 from the vertex point </span>
<span>OVER 3 (right or left) from the vertex point, UP 3² = 9 from the vertex point </span>
<span>OVER 4 (right or left) from the vertex point, UP 4² = 16 from the vertex point </span>
<span>and so on ... </span>
<span>where the "UP" numbers are the sequence of "PERFECT SQUARE" numbers ... </span>
<span>but always counting from the VERTEX EACH time. </span>
<span>➋ another way you can use a "step pattern" is just as you were doing: </span>
<span>Starting with the vertex as "the first point" ... </span>
<span>over 1 (right or left) from the LAST point, up 1 from the LAST point </span>
<span>over 1 (right or left) from the LAST point, up 3 from the LAST point </span>
<span>over 1 (right or left) from the LAST point, up 5 from the LAST point </span>
<span>over 1 (right or left) from the LAST point, up 7 from the LAST point </span>
<span>and so on ... </span>
<span>where the "UP" numbers are the sequence of "ODD" numbers ... </span>
<span>but always counting from the LAST point EACH time. </span>
<span>The reason why both Step Patterns Systems work is that set of PERFECT SQUARE numbers has the feature that the difference between consecutive members is the set of ODD numbers. </span>
<span>For your set of points, the vertex (and all the others) are simply "down 3" from the "standard places": </span>
<span>Standard {..., (-3, 9), (-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4), (3, 9), ...} </span>
<span>shift ↓ 3 : {..., (-3, 6), (-2, 1), (-1, -2), (0, -3), (1,-2), (2, 1), (3, 6), ...} </span>