Answer:
<em>perimeter</em><em> </em><em>of </em><em>figure</em><em> </em><em>A </em><em>is </em><em>4</em><em>0</em><em> </em><em>cm</em>
<em>.</em><em>.</em>
<em>breadth </em><em>of </em><em>figure</em><em> </em><em>B </em><em>is </em><em>3</em><em>.</em><em>3</em><em>3</em><em>. </em><em> </em><em>{</em><em>ie,</em><em>3</em><em> </em><em>whole </em><em>4</em><em> </em><em>by </em><em>1</em><em>2</em><em>}</em>
Answer:
Point of intersections are (0, -7) and (5, -2).
Step-by-step explanation:
From the graph attached,
A straight line is intersecting the circle at the two points (0, 7) and (5, -2).
Now solve algebraically,
Equation of the line → y = x - 7 -------(1)
Equation of the circle → (x - 5)² + (y + 7)² = 25 -------(2)
By substituting the value of y from equation (1) to equation (2)
(x - 5)² + (x - 7 + 7)² = 25
(x - 5)² + x² = 25
x² - 10x + 25 + x² = 25
2x² - 10x = 0
x² - 5x = 0
x(x - 5) = 0
x = 0, 5
From equation (1),
y = 0 - 7 = -7
y = 5 - 7 = -2
Therefore, point of intersections are (0, -7) and (5, -2).
Answer: A
Step-by-step explanation:
I believe you need to solve this using the quadratic formula!
To begin, this is what it is:
x= -b ± <span>√ b^2 - 4ac / 2a
Just plug in what you have in your problem...
2 being a, 13 being b, and -24 being c.
So we get:
x= -13 </span>± <span>√13^2 - 4(2)(-24) / 2(2)
x= -13 </span><span>± √169 - 8 (-24) / 4</span>
<span>x= -13 <span>± √169 + 192 / 4</span>
x= -13 </span>± √<span>361 / 4
The square root of 361 is 19.
So you have: -13 </span><span>± 19 / 4.
Here's where you take the equation </span>-13 <span>± 19 and put the addition and subtraction sign to use.
-13 - 19 = -32
and
-13 + 19 = 6
Now all is left to do is divide the two numbers by 4.
-32/4 = -8
and
6/4 = 3/2
x = -8, 3/2</span>
The principal, real, root of:
![\sqrt[2]{55}](https://tex.z-dn.net/?f=%5Csqrt%5B2%5D%7B55%7D)
=7.41619849
