Ask question

Ask question

All categories

Ray Of Light [21]

3 years ago

12

What are the center and radius of the circle defined by the equation x2 + y2 - 6x+8y +21= 0 ?

1 answer:

solniwko [45]3 years ago

4 0

Answer:

the center is (3, -4) and the radius is = 2

Step-by-step explanation:

You might be interested in

The unbalanced wheel has a mass of 15 kg and rolls without slipping on the flat horizontal surface. The wheel has a radius of gy

miv72 [106K]

Answer:

a) $a_w=4.331m/s^2$

b) $F=53.7N$

c) $\mu=96.5N$

Step-by-step explanation:

From the question we are told that

mass of unbalanced wheel $M=15kg$

Radius of gyration $G=64m$

Mass center $M_c =40mm=$ $4*10^{-2} m$

Angular velocity $\omega=4rads/s$

a)

Generally the equation for moment at point A is mathematically given as

$\sum M_a=I \alpha+\sum M \=ad$

Where $a_w=wheel\ accelaration$

$15*9.81*(0.04)=0.06144*\frac{a_w}{0.1}+15(0.04)(0.04)\frac{a_w}{0.1}+(15a_w-15(0.04)(4^2)0.1)$

$5,87=(a_w*0.8544+15a_0-0.96)$

$a_w=4.331m/s^2$

b)

Generally the equation for equilibrium of system on the horizontal axis is mathematically given as

$\sum f_x=ma_w$

$F=ma_w-m \=r w^2$

$F=(25*4.221)-(15*0.04*4^2)$

$F=63.3-9.6$

$F=53.7N$

c)

Generally the equation for equilibrium of system on the vertical axis is mathematically given as

$\sum f_y=ma_w$

Where

$F=\mu-mg+m \=r \alpha$

$\alpha=\frac{a_0}{0.1}$

$\mu-mg+m \=r \alpha=0$

$\mu-15*9.81+15*0.04*\frac{4.221}{0.1}$

$\mu-96.5N=0$

$\mu=96.5$

3 0

3 years ago

Factor and solve the following equation 2x^2 + x - 21 = 0.

umka21 [38]

Answer:

x = -7/2 and 3

Step-by-step explanation:

Use foil. Solve for x.

8 0

3 years ago

Read 2 more answers

What does 2( 1 3/4)+ 3/5 equal

Dovator [93]

Answer: it equals 7

Step-by-step explanation:

3 0

3 years ago

B) (1 - cos B)(1 + cos B) =<br> 1/cosec^2(B)

faltersainse [42]

Answer:

Proved

Step-by-step explanation:

$(1-cosB)(1+cosB)=\frac{1}{cosec^2B} \\\\Use\\\\ (a+b)(a-b)=a^2-b^2 ,\\so,\\=(1)^2-(cosB)^2\\=1-cos^2B\\\\Now\ , we\ know\ \\sin^2B+cos^2B=1\\\\so,\\sin^2B=1-cos^2B\\so we just replace,\\\\=1-cos^2B=>sin^2B\\and\ since\ sinB=\frac{1}{cosecB} \\\\=sin^2B=>\frac{1}{cosec^2B}\\$

4 0

3 years ago

Which list names the sides of polygon ABCD?

Luba_88 [7]

Answer:

AB, BD, DC, CA

Step-by-step explanation:

i hope it's right let me know

if it's wrong it's my fault

6 0

3 years ago