<img src="https://tex.z-dn.net/?f=-%5Csqrt%7B3%7D%20%2B%20%5Csqrt%7B%28%5Csqrt%7B3%7D-1%29%5E2%20%7D" id="TexFormula1" title="-\

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nalin [4]

2 years ago

sqrt{3} + \sqrt{(\sqrt{3}-1)^2 }" alt="-\sqrt{3} + \sqrt{(\sqrt{3}-1)^2 }" align="absmiddle" class="latex-formula">

Mathematics

2 answers:

RoseWind [281]2 years ago

8 0

Q) $-\sqrt{3} + \sqrt{(\sqrt{3}-1)^2 }$ = ?

→ $-\sqrt{3} + \sqrt{(\sqrt{3}-1)^2 }$

→ $-\sqrt{3} + {\sqrt{3}-1 }$

→ $(-\sqrt{3} + \sqrt{3})-1$

→ -1 is the final answer.

hichkok12 [17]2 years ago

3 0

$- 1$

<h2>hope it helps</h2>

see the attachment for the explanation

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PLEASE HELP!!!!

Lunna [17]

NOTES:

Given a quadratic function in standard format (y = ax² + bx + c), the direction of the parabola is as follows:

if "a" is positive, then opens UP
if "a" is negative, then opens DOWN

Given a quadratic function in standard format (y = ax² + bx + c), the vertex can be found as follows:

the Axis Of Symmetry (x-value) is: x = $\frac{-b}{2a}$
y-value is found by plugging in the AOS for "x" in the equation

****************************************************************************************

1) y = x² + 11x + 24

a = +1 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-11}{2(1)}$ = $-\frac{11}{2}$

y = $(-\frac{11}{2})^{2}$ + 11 $(-\frac{11}{2} )$ + 24 = $-\frac{25}{4}$

vertex $(-\frac{11}{2}$ , $-\frac{25}{4})$ is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-3, 0) and (-8, 0), and y-intercept (0, 24)

******************************************************************************************

2) y = -x² - 6x - 8

a = -1 so the parabola opens DOWN

x = $\frac{-b}{2a}$ = $\frac{-(-6)}{2(-1)}$ = $\frac{6}{-2}$ = -3

y = -(-3)² - 6(-3) - 8 = -9 + 18 - 8 = 1

vertex (-3, 1) is in Quadrant 2 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-4, 0), and y-intercept (0, -8)

******************************************************************************************

3) y = x² - 2x + 3

a = 1 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-(-2)}{2(1)}$ = $\frac{2}{2}$ = 1

y = (1)² - 2(1) + 3 = 1 - 2 + 3 = 2

vertex (1, 2) is in Quadrant 1 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 3), and its mirror image (2, 3). There are no x-intercepts

******************************************************************************************

4) y = x² + 4x + 4

a = 1 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-4}{2(1)}$ = -2

y = (-2)² + 4(-2) + 4 = 4 - 8 + 4 = 0

vertex (-2, 0) is in Quadrant 2 and is on the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 4), and its mirror image (-4, 4). The x-intercept is the vertex.

Compared to the other four graphs, this is most likely the equation for the rain gauge!

******************************************************************************************

5) y = 3x² + 21x + 30

a = +3 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-21}{2(3)}$ = $-\frac{7}{2}$

y = 3 $(-\frac{7}{2})^{2}$ + 21 $(-\frac{7}{2} )$ + 30 = $-\frac{27}{4}$

vertex $(-\frac{7}{2}$ , $-\frac{27}{4})$ is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-5, 0), and y-intercept (0, 30)

*******************************************************************************************

4 0

4 years ago

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Apply the dilation (x, y) = (3x, 3y). What will be the coordinates of T'?

lorasvet [3.4K]

Answer:

6,0

Step-by-step explanation:

the dilation moves points from A to B

in this case we're moving point T

point T has coordinates 2,0 to start with (A)

Now let's move it to B

To do so, we need to apply the calculation

(x,y) => (3x,3y)

Substitute in point T:

(2,0) => (3x,3y)

(2,0) => (3(2),3(0))

(2,0) => (6,0)

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melamori03 [73]

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Arisa [49]

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