Answer:
Let P be the external point. O be the origin. join O and P get OP and nearest point on the circle from P be A.
Let Q be the point onthe circle in which, tangent make 90° with radius at Q.
PQ = 8 and OQ = 6
we get a right angled triangle PQO right angled at Q.
so, OP^2 = OQ^2 + PQ^2= 8^2 + 6^2 = 64 + 36 =1==
therefore OP =10cm
we need nearest point from P, which is PA
PA = OP - OA= 10 -6=4cm
The real solutions the equation as given in the task content; x² = 225 are; +25 and -25.
<h3>What are the real solutions of the equation as given in the task content?</h3>
It follows from the task content that the real solutions of the equation as given in the task content can be determined as follows;
x² = 225
x = ± 15
Therefore, the real solutions of the equation are; +25 and -25.
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Answer:
Option C: 
Step-by-step explanation:
Given the side length of a square is 11 units.
Note that 
Since, length cannot be negative, we can eliminate - 11.
Hence, the length of the side of the square is 11 units = 
.
Hence, the answer.
Answer:
Step-by-step explanation:
Use PEMDAS to solve
(Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction)
Solve parenthesis first
<span>(1/3+9) x ( 8-3)
</span>28/3 x 5
Use Multiplication
28/3 x 5 ≈ 4.7
After rounding 4.7 should be your answer
