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2 years ago

Consider the equation 5x (2x + 1) - 3(2x + 1) = 0.

Mathematics

1 answer:

kondor19780726 [428]2 years ago

7 0

Answer:

It is equivalent to 15x – 3)(2x + 1) = 0. False

It is equivalent to (5x – 3)(2x + 1) = 0 True

Either x=-1/2 or 5/3

Step-by-step explanation:

Consider the equation 5x (2x + 1) - 3(2x + 1) = 0.

It is equivalent to 15x – 3)(2x + 1) = 0. False

It is equivalent to (5x – 3)(2x + 1) = 0 True

When you take pull-out (2x+1), you will be left with (5x-3)

5x (2x + 1) - 3(2x + 1) = 0.

(2x+1)(5x-3)=0

Either 2x + 1 = 0

2x = -1

Divide both-side of the equation by 2

2x/2 = -1/2

x= -1/2

5x-3 =0

5x = 3

Divide both-side of the equation by 5

5x/5 = 3/5

x = 5/3

Either x=-1/2 or 5/3

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The management of an electronics firm has determined that the profit the company earns is a function of thenumber of employees i

ruslelena [56]

we have the function

$P(x)=\ln\left|4x+1\right|+3x-x^2$

For x=2

substitute

$\begin{gathered} P(x)=\operatorname{\ln}\lvert4(2)+1\rvert+3(2)-(2)^2 \\ P(x)=\operatorname{\ln}\lvert9\rvert+2 \\ P(x)=4.20\text{ millions of dollars} \end{gathered}$

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<h2>the answer is 4.20 million dollars</h2>

6 0

7 months ago

Write the equation of a line in point-slope form that goes through the coordinate

Neporo4naja [7]

Answer:

hope this helps you

4 0

2 years ago

While training for a marathon, Jeff wants to increase the number of miles he runs each day. On the first day of training, Jeff r

user100 [1]

Answer:

The series is 5 , 6 , 7 , 8 , 9 , 10 , 11

Step-by-step explanation:

* Lets revise the arithmetic series

- In the arithmetic series there is a constant difference between

each two consecutive numbers

- Ex:

# 2 , 5 , 8 , 11 , ………………………. (constant difference is 3)

# 5 , 10 , 15 , 20 , ………………………… (constant difference is 5)

# 12 , 10 , 8 , 6 , …………………………… (constant difference is -2)

* General term (nth term) of an Arithmetic series:

- If the first term is a and the common diffidence is d, then

U1 = a , U2 = a + d , U3 = a + 2d , U4 = a + 3d , U5 = a + 4d

- So the nth term is Un = a + (n – 1)d, where n is the position of the

number in the series

* Lets solve the problem

- Jeff wants to increase the number of miles he runs each day

∴ He will add the initial value by constant number each day

- He plans on increasing the number of miles he runs a day by 1

∴ The constant value is 1 mile

- On the first day of training, Jeff runs 5 miles

∴ The first value is 5 miles

∴ The series is arithmetic

∵ a = 5 , d = 1

- He do that for the remainder of the week

∵ The week has 7 days

∴ The series has 7 terms

∵ The rule of the series is Un = a + (n - 1)d

∵ a = 5 and d = 1

∴ Un = 5 + (n - 1)(1)

∴ Un = 5 + n - 1

∴ Un = 4 + n ⇒ n is the position of the number

- Substitute n from 1 to 7 to find the series

∴ The series is 5 , 6 , 7 , 8 , 9 , 10 , 11

6 0

3 years ago

Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$

$=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0$

7 0

2 years ago

F(n+1)=f(n)-5. If f(1)=100, what is f(6)

PilotLPTM [1.2K]

Answer:

Step-by-step explanation:

Given:

f(1) = 100

f(n+1) = f(n) - 5

Solve for:

f(6)

Solution:

f(1) = 100

f(2) = f(1) - 5 = 100 - 5 = 100 - 1 x 5 = f(1) - 1 x 5

f(3) = f(2) - 5 = 100 - 5 - 5 = 100 - 2 x 5 = f(1) - 2 x 5

...

f(n) = f(n - 1) - 5 = 100 - (n-1) x 5 = f(1) - (n-1) x 5

=> f(6) = f(1) - (6-1) x 5 = 100 - 5 x 5 = 100 - 25 = 75

Hope this helps!

8 0

2 years ago