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
Answer:
$45
Step-by-step explanation:
Here we need to calculate the income of this year.
We know that a year has 52 weeks. And, our payed weeks are 51, they are, the 50 weeks we work plus the one week of paid-vacation. The remaining week does not give us income, as is unpaid. So our total year income is:
51 * $615 = $31,365
So, our surplus will be our income minus our expenses:
Surplus = $31,365 - $31,320 = $45
Our cash surplus is $45
Answer:
y = 3/7x - 6
General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
- Equality Properties
<u>Algebra I</u>
Slope-Intercept Form: y = mx + b
Step-by-step explanation:
<u>Step 1: Define</u>
Slope <em>m</em> = 3/7
Point (14, 0)
<u>Step 2: Find y-intercept </u><em><u>b</u></em>
- Substitute: 0 = 3/7(14) + b
- Multiply: 0 = 6 + b
- Isolate <em>b</em>: -6 = b
- Rewrite: b = -6
<u>Step 3: Write linear equation</u>
y = 3/7x - 6
Answer:
Step-by-step explanation:
Given equation is,
x² + (p + 1)x = 5 - 2p
x² + (p + 1)x - (5 - 2p) = 0
x² + (p + 1)x + (2p - 5) = 0
Properties for the roots of a quadratic equation,
1). Quadratic equation will have two real roots, discriminant will be greater than zero. [(b² - 4ac) > 0]
2). If the equation has exactly one root, discriminant will be zero [(b² - 4ac) = 0]
3). If equation has imaginary roots, discriminant will be less than zero [(b² - 4ac) < 0].
Discriminant of the given equation = 
For real roots,
p² + 2p + 1 - 8p + 20 > 0
p² - 6p + 21 > 0
For all real values of 'p', given equation will be greater than zero.
