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zavuch27 [327]
2 years ago
15

Find all values of $x$ such that $\sqrt{4x^2} - \sqrt{x^2} = 6$.

Mathematics
1 answer:
Eddi Din [679]2 years ago
4 0

9514 1404 393

Answer:

  x = -6 or 6

Step-by-step explanation:

The square root is always positive, so the equation simplifies to ...

  \sqrt{4x^2}-\sqrt{x^2}=6\\\\2|x|-|x|=6\\\\|x|=6\\\\x=\{-6,\ 6\}

_____

The graph shows the solutions to f(x) = 0, where f(x) = √(4x²) -√(x²) -6.

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¿Cuántos litros hay en 25 vasos si cada vaso contiene 0,20 litros?
Norma-Jean [14]
Voy a decir que son 5 litros
4 0
3 years ago
If u answer ill give brainlist
makkiz [27]

Answer:

120; 643 units squared

Step-by-step explanation:

Perimeter: Add all the sides, so 120 units

_________________________________

Area: First multiply 19 by 24 for the first shape within this other shape, so That equals 456 units squared. Since we know that one whole side is 36 we can subtract 19 from it to get the lower side of the smaller shape and then multiply that by 11. So the smaller side ends up being 17. So 17 times 11 equals: 187 units squared. Then just add the areas together 187+456=643 units squared.

3 0
3 years ago
Which expression are equivalent to -3-(7.5+4)
Hunter-Best [27]
-14.5

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3 0
3 years ago
Which number line shows the solutions to n> 0?
Brilliant_brown [7]

Answer:

Choice A

Step-by-step explanation:

There should be an open circle at 0 since there is no equals in the inequality

n is greater than zero so the line goes to the right.

Choice A

5 0
2 years ago
Read 2 more answers
Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

y'(x)= x^2y(x)-\frac12y^2(x)

Putting the value of y'(x) in equation (1)

y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))

Substituting x =0 and h= 0.2

y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]

\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]    [∵ y(0) =3 ]

\Rightarrow y(0.2)\approx 2.7

Substituting x =0.2 and h= 0.2

y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]

\Rightarrow y(0.4)\approx  2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]

\Rightarrow y(0.4)\approx 1.9926

Substituting x =0.4 and h= 0.2

y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]

\Rightarrow y(0.6)\approx  1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]

\Rightarrow y(0.6)\approx 1.6593

Substituting x =0.6 and h= 0.2

y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]

\Rightarrow y(0.8)\approx  1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]

\Rightarrow y(0.6)\approx 0.8800

Substituting x =0.8 and h= 0.2

y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]

\Rightarrow y(1.0)\approx  0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]

\Rightarrow y(1.0)\approx 0.9152

Therefore the value of y(1)= 0.9152.

4 0
3 years ago
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