Question

1. what are the values of a and b?

$(x + y)^7 = x^7 + 7x^6y + ax^5y^2 + 35x^4y^b + 35x^3y^4 + 21x^2y^5 + 7xy^6 + y^7$
$a = $
$b = $

1. solve.

$|x - 3| = 4$
\\(\bigcirc\\) 7,1
\\(\bigcirc\\) -1,-7
\\(\bigcirc\\) 3,6
\\(\bigcirc\\) 7,-1

1. given the center and radius, write the standard form equation of the circle.

center = (3,4), radius = 2
\\(\bigcirc\\) $(x - 4)^2 + (y - 3)^2 = 4$
\\(\bigcirc\\) $(x + 3)^2 + (y + 4)^2 = 4$
\\(\bigcirc\\) $(x - 3)^2 + (y - 4)^2 = 2$
\\(\bigcirc\\) $(x - 3)^2 + (y - 4)^2 = 4$

Explanation:

Question 8

Step1: Use binomial coefficient formula

The binomial theorem states $(x+y)^n = \sum_{k=0}^n \binom{n}{k}x^{n-k}y^k$, where $\binom{n}{k}=\frac{n!}{k!(n-k)!}$.

Step2: Calculate $a$ (coefficient of $x^5y^2$)

For $x^5y^2$, $n=7$, $k=2$.
$\binom{7}{2}=\frac{7!}{2!(7-2)!}=\frac{7\times6}{2\times1}=21$

Step3: Find $b$ (exponent of $y$)

In the term $35x^4y^b$, $\binom{7}{b}=35$. Since $\binom{7}{3}=\frac{7!}{3!4!}=35$, so $b=3$.

Step1: Split absolute value equation

$|x-3|=4$ splits into two cases:
Case 1: $x-3=4$
Case 2: $x-3=-4$

Step2: Solve Case 1

$x=4+3=7$

Step3: Solve Case 2

$x=-4+3=-1$

Step1: Recall circle standard form

The standard form of a circle is $(x-h)^2+(y-k)^2=r^2$, where $(h,k)$ is the center, $r$ is radius.

Step2: Substitute given values

Center $(3,4)$ so $h=3$, $k=4$; radius $r=2$, so $r^2=4$.
Substitute into formula: $(x-3)^2+(y-4)^2=4$

Answer:

$a=21$, $b=3$

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Question 9