Solve each quadratic equation by factoring. 1] x^2+24=11x 2] 2x^2+12x-14=0

How would I solve this equation for mass (m)?

zaharov [31]

Simple...

you have: $F= \frac{G m^{2} }{ d^{2} }$

You want to find m-->>>

1.)Isolate m

$d^{2} *F= \frac{G m^{2} }{ d^{2} } * d^{2}$

Leaving you with...

F* $d^{2}$ = $G m^{2}$

Keep isolating m-->>>

$F* d^{2} =G m^{2}$

$\frac{F* d^{2} }{G} = \frac{G m^{2} }{G}$

Square root to solve what just m is-->>>

$\sqrt{ \frac{F* d^{2} }{G} } = \sqrt{ m^{2} }$

= $\sqrt{ \frac{F* d^{2} }{G} } =m$

That is how you solve for m...

Thus, your answer.

7 0

3 years ago

In integers how to know if minus (-) 1 upon 2 is greater or minus (-) 1 is greater???

kogti [31]

Answer:

1/2 is 0.5 so -1/2=-0.5

-0.5 is greater than -1

use a number line

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Use distributive property to rewrite 9(10+x)

horsena [70]

Distributive property states a(b+c) = ab + ac

9(10 + x) = 90 + 9x

Your answer is B. 90 + 9x.

5 0

3 years ago

Read 2 more answers

Find the line integral with respect to arc length ∫C(9x+5y)ds, where C is the line segment in the xy-plane with endpoints P=(2,0

Gemiola [76]

(a) You can parameterize C by the vector function

r(t) = (x(t), y(t) ) = P (1 - t ) + Q t = (2 - 2t, 7t )

where 0 ≤ t ≤ 1.

(b) From the above parameterization, we have

r'(t) = (-2, 7) ==> ||r'(t)|| = √((-2)² + 7²) = √53

Then

ds = √53 dt

and the line integral is

$\displaystyle\int_C(9x(t)+5y(t))\,\mathrm ds = \boxed{\sqrt{53}\int_0^1(17t+18)\,\mathrm dt}$

(c) The remaining integral is pretty simple,

$\displaystyle\sqrt{53}\int_0^1(17t+18)\,\mathrm dt = \sqrt{53}\left(\frac{17}2t^2+18t\right)\bigg|_{t=0}^{t=1} = \boxed{\frac{53^{3/2}}2}$

6 0

3 years ago

3x+2(5x-3)=7 with steps

EleoNora [17]

Answer: x=1

Step-by-step explanation: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

3*x+2*(5*x-3)-(7)=0

3x + 2 • (5x - 3)) - 7 = 0

Pull out like factors :

13x - 13 = 13 • (x - 1)

13 = 0

A a non-zero constant never equals zero.

Solve : x-1 = 0

Add 1 to both sides of the equation :

x = 1

6 0

3 years ago