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1 month ago

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Solve (x + 3)2 + (x + 3) – 2 = 0. Let u = Rewrite the equation in terms of u. (u2 + 3) + u – 2 = 0 u2 + u – 2 = 0 (u2 + 9) + u –

2 = 0 u2 + u + 1 = 0 Factor the equation. What are the solutions of the original equation?

2 answers:

lawyer [12.5K]1 month ago

6 0

Answer:

Part 1 gives x + 3

Part 2 yields B u2 + u - 2 + 0

Part 3 shows (u + 2)(u - 1) = 0

Part 4 indicates x = -5 or x = -2

Step-by-step explanation:

on edg... Good Luck!!!

AnnZ [12.3K]1 month ago

4 0

Answer:

The roots of the initial equation are x=-5 and x=-2

Step-by-step explanation:

we have

$(x+3)^2+(x+3)-2=0$

Let

$u=(x+3)$

Rephrase the equation

$(u)^2+(u)-2=0$

Complete the square

$u^2+u=2$

$u^2+u+1/4=2+1/4$

$u^2+u+1/4=9/4$

rewrite as complete squares

$(u+1/2)^2=9/4$

take the square root of both sides

$(u+1/2)=\pm\frac{3}{2}$

$u=(-1/2)\pm\frac{3}{2}$

$u=(-1/2)+\frac{3}{2}=1$

$u=(-1/2)-\frac{3}{2}=-2$

the solutions are

u=-2,u=1

Alternative Method

The equation used to solve a quadratic of the type

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this case we have

$(u)^2+(u)-2=0$

so

$a=1\\b=1\\c=-2$

insert into the formula

$u=\frac{-1\pm\sqrt{1^{2}-4(1)(-2)}} {2(1)}$

$u=\frac{-1\pm\sqrt{9}} {2}$

$u=\frac{-1\pm3} {2}$

$u=\frac{-1+3} {2}=1$

$u=\frac{-1-3} {2}=-2$

the discovered roots are

u=-2,u=1

Find the roots of the initial equation

For u=-2

$-2=(x+3)$ ----> $x=-2-3=-5$

For u=1

$1=(x+3)$ ----> $x=1-3=-2$

therefore

The roots of the initial equation are

x=-5 and x=-2

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A plane ticket barcelona costs £175 the price decreases by 6% work out the new price

Svet_ta [12734]

Answer: £164.50

Reducing 175 by 6% results in 164.5.

The absolute change is:

164.5 - 175 = -10.5

Step-by-step explanation:

The calculation is as follows:

175 - Percentage decrease =

175 - (6% × 175) =

175 - 6% of 175 =

(1 - 0.06) × 175 =

0.94 × 175 =

94 ÷ 100 × 175 =

94 × 175 divided by 100 =

16,450 ÷ 100 =

164.5

So, the final amount is £164.50

6 0

1 month ago

Read 2 more answers

Evaluate 0.1m+8-12n0.1m+8−12n0, point, 1, m, plus, 8, minus, 12, n when m=30m=30m, equals, 30 and n=\dfrac14n= 4 1 n, equals,

PIT_PIT [12445]

Answer:

8

Step-by-step explanation:

The task is to evaluate:

0.1m + 8 - 12n

When $m=30, n=\frac{1}{4}$

By substituting these values into the expression, we have:

$0.1m+8-12n=0.1(30)+8-12(\frac{1}{4})\\=3+8-\frac{12}{4}\\=3+8-3\\=8$

8 0

11 days ago

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a resorvoir can be filled by an inlet pipe in 24 hours and emptied by an outlet pipe 28 hours. the foreman starts to fill the re

zzz [12365]

To determine the rates at which the inlet and outlet pipes fill and empty the reservoir, we remember that work done equals rate multiplied by time. Let’s denote the inlet rate as i and for the outlet pipe as 0. Therefore,
i(24) = 1
o(28) = 1
In this context, the '1' represents the total number of reservoirs, since the problem states the time needed for each pipe to either fill or empty a singular reservoir. Solving for rates yields:
i = 1/24 reservoirs/hour
o = 1/28 reservoirs/hour

Over the first six hours, the inlet pipe fills (1/24)(6) = 1/4 reservoirs and during the same period, the outlet pipe empties (1/28)(6) = 3/14 reservoirs. To calculate the net volume of the reservoir filled, we subtract the emptying total from the filling total:
1/4 - 3/14 = 1/28 reservoirs (note that if emptying exceeds filling, a negative value results. In such cases, treat that negative value as zero, indicating that the outlet rate surpasses the inlet rate, leading to an empty reservoir).
Now we need to find out how long it will take to fill up one reservoir since we’ve already partially filled 1/28 of it, after closing the outlet pipe. In simpler terms, we need to determine the time required for the inlet pipe to finish filling the remaining 27/28 of the reservoir. Fortunately, we have already established the filling rate for the inlet pipe, leading to the equation:
(1/24)t = 27/28
Solving for t gives us 23.14 hours. Remember to add the initial 6 hours to this result since the question seeks the total time. Thus, the final total is 29.14 hours.

Please ask me any questions you may have!

4 0

1 month ago

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A ship anchored in a port has a ladder which hangs over the side. The length of the ladder is 200cm, the distance between each r

Leona [12618]

Response:

The answer to the inquiry is 8 hours.

Step-by-step breakdown:

Information

Ladder length = 200 cm

Distance between rungs = 20 cm

Tide rise rate = 10 cm/h

Fifth rung =?

Procedure

1.- Determine the total height the tide must reach

Height = 20 cm x 4

= 80 cm as the first rung is touching the water.

2.- Calculate the time needed

Rate = distance / time

-Solve for time

Time = distance / rate

-Substitute values

Time = 80 cm / 10 cm/h

-Final outcome

Time = 8 hours.

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1 month ago

The ground-state wave function for a particle confined to a one-dimensional box of length L is Ψ=(2/L)^1/2 Sin(πx/L). Suppose th

AnnZ [12381]

Respuesta:

(a) 4.98x10⁻⁵

(b) 7.89x10⁻⁶

(c) 1.89x10⁻⁴

(d) 0.5

(e) 2.9x10⁻²

Explicación paso a paso:

La probabilidad (P) de encontrar la partícula está dada por:

$P=\int_{x_{1}}^{x_{2}}(\Psi\cdot \Psi) dx = \int_{x_{1}}^{x_{2}} ((2/L)^{1/2} Sin(\pi x/L))^{2}dx$

$P = \int_{x_{1}}^{x_{2}} (2/L) Sin^{2}(\pi x/L)dx$ (1)

La solución de la integral de la ecuación (1) es:

$P=\frac{2}{L} [\frac{X}{2} - \frac{Sin(2\pi x/L)}{4\pi /L}]|_{x_{1}}^{x_{2}}$

(a) La probabilidad de encontrar la partícula entre x = 4.95 nm y 5.05 nm es:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{4.95}^{5.05} = 4.98 \cdot 10^{-5}$

(b) La probabilidad de encontrar la partícula entre x = 1.95 nm y 2.05 nm es:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{1.95}^{2.05} = 7.89 \cdot 10^{-6}$

(c) La probabilidad de encontrar la partícula entre x = 9.90 nm y 10.00 nm es:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{9.90}^{10.00} = 1.89 \cdot 10^{-4}$

(d) La probabilidad de encontrar la partícula en la mitad derecha de la caja, es decir, entre x = 0 nm y 50 nm es:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{50.00} = 0.5$

(e) La probabilidad de encontrar la partícula en el tercio central de la caja, es decir, entre x = 0 nm y 100/6 nm es:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{16.7} = 2.9 \cdot 10^{-2}$

Espero que te ayude.

3 0

25 days ago