Answer:
1. 24.36
2. Do the same thing (follow the steps in the picture)
Step-by-step explanation:
Two circles<span> of </span>radius<span> 4 are </span>tangent<span> to the </span>graph<span> of y^</span>2<span> = </span>4x<span> at the </span>point<span> (</span>1<span>, </span>2<span>). ... I know how to </span>find<span> the </span>tangent<span> line from a circle and a given </span>point<span>, but ... </span>2a2=42. a2=8. a=±2√2. Then1−xc=±2√2<span> and </span>2−yc=±2√2. ... 4 from (1,2<span>), so you could </span>find these<span> centers, and from there the</span>equations<span> of the circle
</span>
Answer:
y = 2x - 7
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
y = 2x + 3 ← is in slope- intercept form
with slope m = 2
Parallel lines have equal slopes , then
y = 2x + c ← is the partial equation
To find c substitute (5, 3 ) into the partial equation
3 = 10 + c ⇒ c = 3 - 10 = - 7
y = 2x - 7 ← equation of parallel line
Answer:
Below.
Step-by-step explanation:
f) (a + b)^3 - 4(a + b)^2
The (a+ b)^2 can be taken out to give:
= (a + b)^2(a + b - 4)
= (a + b)(a + b)(a + b - 4).
g) 3x(x - y) - 6(-x + y)
= 3x( x - y) + 6(x - y)
= (3x + 6)(x - y)
= 3(x + 2)(x - y).
h) (6a - 5b)(c - d) + (3a + 4b)(d - c)
= (6a - 5b)(c - d) + (-3a - 4b)(c - d)
= -(c - d)(6a - 5b)(3a + 4b).
i) -3d(-9a - 2b) + 2c (9a + 2b)
= 3d(9a + 2b) + 2c (9a + 2b)
= 3d(9a + 2b) + 2c (9a + 2b).
= (3d + 2c)(9a + 2b).
j) a^2b^3(2a + 1) - 6ab^2(-1 - 2a)
= a^2b^3(2a + 1) + 6ab^2(2a + 1)
= (2a + 1)( a^2b^3 + 6ab^2)
The GCF of a^2b^3 and 6ab^2 is ab^2, so we have:
(2a + 1)ab^2(ab + 6)
= ab^2(ab + 6)(2a + 1).
Answer:
80
Step-by-step explanation:
The Two-Tangent Theorem states that if two tangent segments are drawn to one circle from the same external point, then they are congruent
13+13+9+9+12+12+6+6= 80
