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4 years ago

Solve for x show all work -2× + 5=4× -7

Mathematics

2 answers:

Artyom0805 [142]4 years ago

8 0

-2x+5=4x-7

5=6x-7

13=6x

2.5=x

Anestetic [448]4 years ago

6 0

5=4x+2x-7
5+7=6x
12=6x
12/6=6x/6
2=x

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3 years ago

Solve the equation 2 sec² x = 3 - tan x for the domain 0° ≤ x ≤ 360°

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$2 \sec^2 x = 3- \tan x \\\\\implies 2(1 +\tan^2 x) = 3- \tan x \\\\\implies 2 + 2 \tan^2 x +\tan x -3 =0\\\\\implies 2 \tan^2 x + \tan x -1 =0\\\\\implies 2u^2 + u -1 =0~~~ ;[\text{set} \tan x = u]\\\\\implies u = \dfrac{-1\pm \sqrt{1-4\cdot 2 \cdot (-1)}}{2(2)}\\\\\implies u =\dfrac{-1 \pm\sqrt{9}}{4}\\\\\implies u = \dfrac{-1 \pm 3}{4}\\\\\implies u = \dfrac{2}{4}=\dfrac 12~~ \text{or} ~~u =\dfrac{-4}{4} =-1\\\\$

$\implies \tan x = \dfrac 12 ~~ \text{or} ~~ \tan x =-1~~~ ;[\text{Substitute back u =tan x}]\\\\\text{Now,}~ \\\\\tan x = -1,\\\\\implies x = n\pi - \dfrac{\pi}4\\\\\\\text{For interval,}~ [0,2\pi) \\\\x = \dfrac{3\pi}4, \dfrac{7\pi}4\\\\\text{In degrees,}~ x = 135^{\circ}, x =315^{\circ}\\\\\tan x = \dfrac 12\\\\\implies x = n\pi + \tan^{-1} \left(\dfrac 12 \right)\\\\\text{For interval} ~[0,2\pi),\\\\$

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4 0

2 years ago

Plis i need help whit this questionss

zmey [24]

Question 1: x+2 = 8 doesn't belong in the given equations.

Question 2: The possible values for y are 0,1 and 2

Step-by-step explanation:

Question: Which one doesn't belong?

We will solve each equation one by one to get which equation doesn't belong with others

So,

$x+4 = -2\\x+4-4 = -2-4\\x = -6$

$x+5 = -1\\x+5-5 = -1-5\\x = -6$

$x+2 = 8\\x+2-2 = 8-2\\x = 6$

$3-x=9\\3-x-3 = 9-3\\-x = 6\\x = -6$

Hence,

x+2 = 8 doesn't belong in the given equations.

Question 2:

Given equation is:

$x+y = 5$

It is given that the values of x will be greater than 2 and less than 6 so values of x can be 3,4 and 5

So,

$3+y = 5\\y = 5-3\\y = 2\\\\4+y = 5\\y = 5-4\\y = 1\\\\5+y = 5\\y = 5-5\\y = 0$

Hence,

The possible values for y are 0,1 and 2

So,

x+2 = 8 doesn't belong in the given equations.

The possible values for y are 0,1 and 2

Keywords: Linear equation, variables

Learn more about linear equations at:

#LearnwithBrainly

8 0

3 years ago

The coordinate plane below represents a city. Points A through F are schools in the city. graph of coordinate plane. Point A is

Ray Of Light [21]

Part A;
There are many system of inequalities that can be created such that only contain points A and E in the overlapping shaded regions.

Any system of inequalities which is satisfied by (2, -3) and (3, 1) but is not satisfied by (-3, -4), (-4, 2), (2, 4) and (-2, 3) can serve.

An example of such system of equation is
y ≤ x
y ≥ -2x
The system of equation above represent all the points in the first quadrant of the coordinate system.
The area above the line y = -2x and to the right of the line y = x is shaded.

Part B:
It can be verified that points A and E are solutions to the system of inequalities above by substituting the coordinates of points A and E into the system of equations and see whether they are true.

Substituting A(2, -3) into the system we have:
-3 ≤ 2
-3 ≥ -2(2) ⇒ -3 ≥ -4
as can be seen the two inequalities above are true, hence point A is a solution to the set of inequalities.

Also, substituting E(3, 1) into the system we have:
1 ≤ 3
1 ≥ -2(3) ⇒ 1 ≥ -6
as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.

Part C:
Given that William can only attend a school in her designated zone and that William's zone is defined by y < −x - 1.

To identify the schools that William is allowed to attend, we substitute the coordinates of the points A to F into the inequality defining William's zone.

For point A(2, -3): -3 < -(2) - 1 ⇒ -3 < -2 - 1 ⇒ -3 < -3 which is false

For point B(-3, -4): -4 < -(-3) - 1 ⇒ -4 < 3 - 1 ⇒ -4 < 2 which is true

For point C(-4, 2): 2 < -(-4) - 1 ⇒ 2 < 4 - 1 ⇒ 2 < 3 which is true

For point D(2, 4): 4 < -(2) - 1 ⇒ 4 < -2 - 1 ⇒ 4 < -3 which is false

For point E(3, 1): 1 < -(3) - 1 ⇒ 1 < -3 - 1 ⇒ 1 < -4 which is false

For point F(-2, 3): 3 < -(-2) - 1 ⇒ 3 < 2 - 1 ⇒ 3 < 1 which is false

Therefore, the schools that Natalie is allowed to attend are the schools at point B and C.

4 0

3 years ago

Solve for x show all work -2× + 5=4× -7​

Solve for x show all work -2× + 5=4× -7