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2 years ago

Find the condition of the expression:√3-2x-x^2

Mathematics

1 answer:

Alex73 [517]2 years ago

7 0

Answer:

-1 <= x <= 1

Step-by-step explanation:

√((1-2x+x^2)+2-2x^2)

= √((1-x)^2 + 2*(1-x)(1+x) )

điều kiện là 1 - x > =0 và 1+x >=0

giải 2 bất phương trình trên ta thu đc kết ququả

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: Which equations have an infinite number of solutions? Select all that apply. A -2(x - 2) + 3x = x - 4 B 2x + 4(x - 1) = 3(2x +

MissTica

Option C and Option D

3x + 2(x - 2) + 6 = 2(2x + 3) + x - 4 has infinite number of solutions

4(x + 2) + x + 1 = 2x - 3 + 3(x + 4) has infinite number of solutions

<h3><u>Solution:</u></h3>

Given that we have to find the equations that has infinite number of solutions

If we end up with the same term on both sides of the equal sign, such as 4 = 4 or 4x = 4x, then we have infinite solutions.

If we end up with different numbers on either side of the equal sign, as in 4 = 5, then we have no solutions.

<h3><u>Option A</u></h3>

-2(x - 2) + 3x = x - 4

Multiply the terms inside the bracket

-2x + 4 + 3x = x - 4

x + 4 = x - 4

Thus this equation does not have infinite number of solutions

<h3><u>Option B</u></h3>

2x + 4(x - 1) = 3(2x + 1) - 2(x - 1)

2x + 4x - 4 = 6x + 3 - 2x + 2

6x - 4 = 4x + 5

6x - 4x = 5 + 4

2x = 9

$x = \frac{9}{2}$

Thus this equation has only one solution.

Thus this equation does not have infinite number of solutions

<h3><u>Option C</u></h3>

3x + 2(x - 2) + 6 = 2(2x + 3) + x - 4

3x + 2x - 4 + 6 = 4x + 6 + x - 4

5x + 2 = 5x + 2

Thus this equation has infinite number of solutions

<h3><u>Option D</u></h3>

4(x + 2) + x + 1 = 2x - 3 + 3(x + 4)

4x + 8 + x + 1 = 2x - 3 + 3x + 12

5x + 9 = 5x + 9

Thus this equation has infinite number of solutions

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3 years ago

Ronald had to drain 5,750 gallons of water out of his pool. Each day he drained 10 times fewer gallons of water than he did the

kenny6666 [7]

Answer:

Step-by-step explanation:

5,750

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3 years ago

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<span>B) (1, 5) is your best answer

B (0,6) creates the top of the kite
D (0,2) creates the bottom
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Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :

$\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}$

$k \approx \pm 0.028$

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I would say for every 1 brownie, there are 5 cookies.

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