80m15 + 72m12 − 32m10

X^2+4x+y^2-10y+20=30 find the center of the circle by completing the square

swat32

Answer:

a). Center of the circle = (-2, 5)

b). Equation of the line ⇒ y = $-\frac{4}{5}x+\frac{58}{5}$

Step-by-step explanation:

Equation of the circle is,

x² + 4x + y²- 10y + 20 = 30

a). [x² + 2(2)x + 4 - 4] + [y²- 2(5)y + 25] - 25 + 20 = 30

[x² + 2(2)x + 4] - 4 + [y² - 2(5)y + 25] - 25 + 20 = 30

(x + 2)² + (y - 5)²- 29 + 20 = 30

(x + 2)² + (y - 5)²- 9 = 30

(x + 2)² + (y - 5)² = 39

By comparing this equation with the standard equation of a circle,

Center of the circle is (-2, 5).

b). A point (2, 10) lies on this circle.

Slope of the line joining this point to the center (-2, 5),

$m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

= $\frac{10-5}{2+2}$

= $\frac{5}{4}$

Let the slope of the tangent which is perpendicular to this line is ' $m_{2}$ '

Then by the property of perpendicular lines,

$m_{1}\times m_{2}=-1$

$\frac{5}{4}\times m_{2}=-1$

$m_{2}=-\frac{4}{5}$

Now the equation of the line passing though (2, 10) having slope $m_{2}=-\frac{4}{5}$

y - y' = $m_{2}(x-x')$

y - 10 = $-\frac{4}{5}(x-2)$

y - 10 = $-\frac{4}{5}x+\frac{8}{5}$

y = $-\frac{4}{5}x+\frac{8}{5}+10$

y = $-\frac{4}{5}x+\frac{58}{5}$

Therefore, equation of the line will be, y = $-\frac{4}{5}x+\frac{58}{5}$

7 0

3 years ago

A rock is thrown upward from a bridge that is 57 feet above a road. The rock reaches its maximum height above the road 0.76 seco

bekas [8.4K]

answer:

$f(t) = -9.9788169(t -0.76)^2 +57$

Step-by-step explanation:

On this question we see that we are given two points on a certain graph that has a maximum point at 57 feet and in 0.76 seconds after it is thrown, we know can say this point is a turning point of a graph of the rock that is thrown as we are told that the function f determines the rocks height above the road (in feet) in terms of the number of seconds t since the rock was thrown therefore this turning point coordinate can be written as (0.76, 57) as we are told the height represents y and x is represented by time in seconds. We are further given another point on the graph where the height is now 0 feet on the road then at this point its after 3.15 seconds in which the rock is thrown in therefore this coordinate is (3.15,0).

now we know if a rock is thrown it moves in a shape of a parabola which we see this equation is quadratic. Now we will use the turning point equation for a quadratic equation to get a equation for the height which the format is $f(x)= a(x-p)^2 +q$ , where (p,q) is the turning point. now we substitute the turning point

$f(t) = a(t-0.76)^2 + 57$ , now we will substitute the other point on the graph or on the function that we found which is (3.15, 0) then solve for a.

0 = a(3.15 - 0.76)^2 + 57

-57 =a(2.39)^2

-57 = a(5.7121)

-57/5.7121 =a

-9.9788169 = a then we substitute a to get the quadratic equation therefore f is

$f(t) = -9.9788169(t-0.76)^2 +57$

4 0

3 years ago

Is it possible for a triangle to have sides with the given lengths: 20 m, 22 m, and 24 m? A. Yes; the sum of each pair is greate

snow_lady [41]

The correct answer is A) yes, the sum of each pair is greater than the third.

The triangle inequality theorem states that the sum of any two sides of a triangle is longer than the third side. This is true for these numbers, so this does form a triangle.

3 0

3 years ago

Read 2 more answers

A King in ancient times agreed to reward the inventor of chess with one grain of wheat on the first of the 64 squares of a chess

maxonik [38]

Answer:

Total weight of the wheat placed on 45th square will be = $1.26\times 10^{6}tons$

Step-by-step explanation:

King rewarded the inventor of chess by giving grain of wheat on every square of the chess board. He places one grain of wheat on first square, 2 on second, 4 on 3rd, 8 on 4th and so on.

In fact he places grains of wheat in a progression S = 1, 2, 4, 8, 16.........n terms where value of n is 64.

We can rewrite the progression as S = 2°, 2, 2², 2³,.........n terms (n = 64)

In a geometric progression we know the nth term = $a(r^{n-1})$