All categories

3 years ago

A^2+2ab+b^2. solve for a=10 and b=50

1 answer:

3 0

A² + 2ab + b² ; Solve when a=10 & b=50

Simply plug in 10 for a and 50 for b in our given equation :)

(10)² + 2(10)(50) + (50)²

Simplify.

100 + 20(50) + 2500

Simplify.

(100 + 2500) + 1000

Simplify.

2600 + 1000

Simplify.

3600

Hope I helped! :)

You might be interested in

Triangle ABC has vertices at A(2,3),B(-4,-3) and C(2,-3) find the coordinates of each point of concurrency.

dem82 [27]

Answer:

Circumcenter =(-1,0)

Orthocenter =(2,-3)

Step-by-step explanation:

Given : Points A = (2,3), B = (-4,-3), C = (2,-3)

Formula used :

→Mid point of two points- $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

→Slope of two points - $\frac{y_2-y_1}{x_2-x_1})$

→Perpendicular of a line = $\frac{-1}{slope of line})$

Circumcenter- The point where the perpendicular bisectors of a triangle meets.

Orthocenter-The intersecting point for all the altitudes of the triangle.

To find out the circumcenter we have to solve any two bisector equations.

We solve for line AB and AC

So, mid point of AB = $(\frac{2-4}{2},\frac{3-3}{2})=(-1,0)$

Slope of AB = $\frac{-3-3}{-4-2}=1$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of the perpendicular bisector = -1

Equation of AB with slope -1 and the coordinates (-1,0) is,

(y – 0) = -1(x – (-1))

y+x=-1………………(1)

Similarly, for AC

Mid point of AC = $(\frac{2+2}{2},\frac{3-3}{2})=(2,0)$

Slope of AC = $\frac{-3-3}{2-2}=\frac{-6}{0}$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of the perpendicular bisector = 0

Equation of AC with slope 0 and the coordinates (2,0) is,

(y – 0) = 0(x – 2)

y=0 ………………(2)

By solving equation (1) and (2),

put y=0 in equation (1)

y+x=-1

0+x=-1

⇒x=-1

So the circumcenter(P)= (-1,0)

To find the orthocenter we solve the intersections of altitudes.

We solve for line AB and BC

So, mid point of AB = $(\frac{2-4}{2},\frac{3-3}{2})=(-1,0)$

Slope of AB = $\frac{-3-3}{-4-2}=1$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of CF = -1

Equation of AB with slope -1 and the coordinates (-1,0) gives equation CF

(y – 0) = -1(x – (-1))

y+x=-1………………(3)

Similarly, mid point of BC = $(\frac{-4+2}{2},\frac{-3-3}{2})=(-1,-3)$

Slope of AB = $\frac{-3+3}{-4-2}=0$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of AD = 0

Equation of AB with slope 0 and the coordinates (-1,-3) gives equation AD

(y-(-3)) = 0(x – (-1))

y+3=0

y=-3………………(4)

Solve equation (3) and (4),

Put y=-3 in equation (3)

y+x=-1

-3+x=-1

x=2

Therefore, orthocenter(O)= (2,-3)

7 0

3 years ago

What is the y-value of the vertex of the function f(x)=-(x-3)(x+11)

lutik1710 [3]

so, this is a quadratic equation, meaning two solutions, and we have a factored form of it, meaning you can get the solutions by simply zeroing out the f(x).

$\bf \stackrel{f(x)}{0}=-(x-3)(x+11)\implies 0=(x-3)(x+11)\implies x= \begin{cases} 3\\ -11 \end{cases} \\\\\\ \boxed{-11}\stackrel{\textit{\large 7 units}}{\rule[0.35em]{10em}{0.25pt}}-4\stackrel{\textit{\large 7 units}}{\rule[0.35em]{10em}{0.25pt}}\boxed{3}$

so the zeros/solutions are at x = 3 and x = -11, now, bearing in mind the vertex will be half-way between those two, checking the number line, that midpoint will be at x = -4, so the vertex is right there, well, what's f(x) when x = -4?

$\bf f(-4)=-(-4-3)(-4+11)\implies f(-4)=7(7)\implies f(-4)=49 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{vertex}{(-4~~,~~49)}~\hfill$