All categories

2 years ago

Solve for m and show work please

Mathematics

1 answer:

strojnjashka [21]2 years ago

4 0

if im correct the answer to that is undefined.

1. subtract 4 from both sides.

9-5m=6m

2. then add 5m to both sides.

9=11m

thats the last step

You might be interested in

A) Expand and simplify (x + 4)(x-7)<br>

saveliy_v [14]

Answer:

Expanded:x²-7x+4x-28

Simplified: x²-3x-28

Step-by-step explanation:

7 0

3 years ago

Please me with my math

Katyanochek1 [597]

Answer:

8/9, $\sqrt{2}$ , 11/5, $\sqrt{9\\}$ , π, 11/3, 5

Step-by-step explanation:

Simplify and this should be the answer

7 0

3 years ago

Read 2 more answers

Solve using the division algorithm 57.27 divided by 23

GenaCL600 [577]

Answer:

2.49

Step-by-step explanation:

57.27 / 23 = 2.49

5 0

3 years ago

Solve the equation using the Properties of Equality.

Simora [160]

Answer:

7a + 3 -3 =45 - 3

7a + 0 = 42

7a = 42

7a/7 = 42/7

a = 6

4 0

2 years ago

In this exercise, consider a particle moving on a circular path of radius b described by r(t) = b cos(ωt)i + b sin(ωt)j, where ω

Bingel [31]

Answer:

Acceleration of the particle = $bw^{2}$

Step-by-step explanation:

We are given the position vector of a particle moving in a circle of radius b units.

r(t) = b cos(ωt)i + b sin(ωt)j

Velocity , v = $\frac{dr}{dt}$ = -bω sin(ωt)i + bω cos(ωt)j

The magnitude of velocity, v = $\sqrt{v_x^{2} +v_y^{2} }$

Squaring both sides,

$v^{2} = b^{2} w^{2}(sin^{2}(wt)+cos^{2}(wt))$

Since $sin^{2}(wt)+cos^{2}(wt))$ = 1

$v^{2} = b^{2}w^{2}$

The acceleration towards the centre is called the centripetal acceleration and is given by

a = $\frac{v^{2} }{r}$

a = $\frac{b^{2}w^{2}}{b}$

a = $bw^{2}$

3 0

2 years ago