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Mademuasel [1]
3 years ago
8

Solve this equation using the radical equation method

Mathematics
1 answer:
Yuki888 [10]3 years ago
6 0
The domain:
3x+6\geq0\ \ |-6\\3x\geq-6\ \ \ |:3\\x\geq-2\\\\D:x\in\left< -2;\ \infty\right)

2x=\sqrt{3x+6}-1\ \ \ |+1\\\\2x+1=\sqrt{3x+6}\ \ \ \ |^2\\\\(2x+1)^2=\left(\sqrt{3x+6}\right)^2\ \ \ |use:\ (a+b)^2=a^2+2ab+b^2\\\\(2x)^2+2\cdot2x\cdot1+1^2=3x+6\\\\4x^2+4x+1=3x+6\ \ \ \ |-3x-6\\\\4x^2+x-5=0\\\\4x^2-4x+5x-5=0\\\\4x(x-1)+5(x-1)=0\\\\(x-1)(4x+5)=0\iff x-1=0\ \vee\ 4x+5=0\\\\x=1\ \vee\ 4x=-5\\\\x=1\in D\ \vee\ x=-1.25\in D

Answer:\ \boxed{x=-1.25\ \vee\ x=-1}
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Setler79 [48]

Answer:

65 square units

Step-by-step explanation:

The area of a triangle is: height times base divided by two.

So the area of one of the triangles is 10 square units.

multiply this times 4 to get the total area of the triangles

then add the area of the square which is 5 times 5

25 plus 40 = 65 square units

PLZ MARK BRAINLIEST

3 0
3 years ago
Read 2 more answers
One of the equations below is not correct. Which on is incorrect. Then explain the error that was made and write the correct ans
timurjin [86]
Only 3 is false, because of spare minus.
3 0
3 years ago
(I need the answer right now please) ΔA'B'C' was constructed using ΔABC and line segment EH. 2 triangles are shown. Line E H is
irina1246 [14]

Answer:

The true statements are

1. BD = DB'

3. m∠EFA = 90°

4. The line of reflection, EH, is the perpendicular bisector of BB', AA', and

   CC'

Step-by-step explanation:

* Lets explain how to solve the problem

- Reflection is flipping an object over the line of reflection.

- The object and its image have the same shape and size, but the

  figures are in opposite directions from the line of reflection

- The objects appear as if they are mirror reflections, with right and left

  reversed

- The line of reflection is a perpendicular bisector for all lines joining

  points on the figure with their corresponding images

- Look to the attached figure for more understand

* Lets solve the problem

- ΔA'B'C' was constructed using ΔABC and line segment EH, where

 EH is the reflection line

- D is the mid-point of BB'

- F is the mid-point of AA'

- G is the mid-point of CC'

* Lets find from the answer the true statements

1. BD = DB'

∵ D is the mid point of BB'

- Point D divides BB' into two equal parts

∴ BD = DB' ⇒ <em>True</em>

2. DF = FG

- It depends on the size of the sides and angles of the triangle

∵ We can't prove that

∴ DF = FG ⇒ <em>Not true</em>

3. m∠EFA = 90°

∵ The line of reflection ⊥ the lines joining the points with their

   corresponding images

∴ EH ⊥ AA' and bisect it at F

∴ m∠EFA = 90° ⇒ <em>True</em>

4. The line of reflection, EH, is the perpendicular bisector of BB',

   AA', and CC'

- Yes the line of reflection is perpendicular bisectors of them

∴ The line of reflection, EH, is the perpendicular bisector of BB',

   AA', and CC' ⇒ <em>True</em>

5. ΔABC is not congruent to ΔA'B'C'

∵ In reflection the object and its image have the same shape and size

∴ Δ ABC is congruent to Δ A'B'C'

∴ ΔABC is not congruent to ΔA'B'C' ⇒ <em>Not true</em>

4 0
3 years ago
Read 2 more answers
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forsale [732]

Answer: 132.27

Step-by-step explanation:

Multiply 8 × 9 × 27 to get the value of x

= 1944

To get 3 square root of 1944, find the square root of 1944 = 44.09

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6 0
3 years ago
What is the radius of a circle if it's area is 36π?
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Area= pi x r^2
hence,

pi x r^2 = 36 x pi

=> r^2=36
=>r= root over 36 = 6

in short...the radius is 6
5 0
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