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2 years ago

In the trapezoid at the right, BE= 2x-8, DE= x-4, and AC= x+2

Mathematics

1 answer:

Marta_Voda [28]2 years ago

5 0

Answer:

A. x = 5 (Equation below)

B. AC = 9 | BD = 9

Step-by-step explanation:

A. Since we know the diagonals of an isosceles trapezoid are congruent, we know that the two bottom parts of the diagonal BD (2x - 8 and x - 4) are congruent to the diagonal AC (x + 2) . This means you can just equal them together and get the value of x:

2x - 8 + x - 4 = x + 2

3x - 12 = x + 2

3x = x + 14

2x = 14

x = 7

B. Just plug in 5 for x:

AC = x + 2 = 7 + 2 = 9

BD = 2x - 8 + x -4 = 3x - 12 = 21 - 12 = 9

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The general form of the equation of a circle is x2 + y2 + 42x + 38y − 47 = 0. The equation of this circle in standard form is .

Readme [11.4K]

Answer:

(-21,-19)

$\sqrt{849}$

Standard form

Step-by-step explanation:

We are given the equation of circle

$x^2+y^2+42x+38y-47=0$

General equation of circle:

$x^2+y^2+2gx+2fy+c=0$

Centre: (-g,-f)

Radius: $\sqrt{g^2+f^2-c}$

Compare the equation to find f, g and c from the equation

$g\rightarrow 21$

$f\rightarrow 19$

$c\rightarrow -47$

Centre: (-21,-19)

Radius (r) $=\sqrt{21^2+19^2+47}=\sqrt{849}$

Standard form of circle:

$(x+21)^2+(y+19)^2=849$

The centre of circle at the point (-21,-19) and its radius is $\sqrt{849}$ .

The general form of the equation of a circle that has the same radius as the above circle is standard form.

4 0

2 years ago

Read 2 more answers

The spring has a stiffness k=200n/m and is unstretched when the 25 kg block is at

g100num [7]

To solve this we are going to use the formula fro the force applied to a spring: $F=ks$
where
$k$ is the spring constant
$s$ is the extension

Since we know the $F=ma$ , we can replace that in our formula and solve for $a$ :
$ma=ks$
$a= \frac{ks}{m}$
where
$a$ is the acceleration
$k$ is the spring constant
$s$ is the extension
$m$ is the mass

We know for our problem that $k=200$ , $s=0.4$ , and $m=25$ . So lets replace those values in our formula to find $a$ :
$a= \frac{ks}{m}$
$a= \frac{(200)(0.4)}{25}$
$a=3.5 \frac{m}{s^2}$

We can conclude that the acceleration of the block when s=0.4m is $3.5 \frac{m}{s^2}$ .