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3 years ago

Help plz plz plz plz

Mathematics

1 answer:

Goshia [24]3 years ago

8 0

Answer:

2.8 In is the answer to the question

Step-by-step explanation:

Nw+WD=ND

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I NEED HELP PLS THIS IS DUE IN 3 HOURS

Mariulka [41]

Answer:

Part 1) $x^{2} -2x-2=(x-1-\sqrt{3})(x-1+\sqrt{3})$

Part 2) $x^{2} -6x+4=(x-3-\sqrt{5})(x-3+\sqrt{5})$

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

Part 1)

in this problem we have

$x^{2} -2x-2=0$

$a=1\\b=-2\\c=-2$

substitute in the formula

$x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(1)(-2)}} {2(1)}\\\\x=\frac{2(+/-)\sqrt{12}} {2}\\\\x=\frac{2(+/-)2\sqrt{3}} {2}\\\\x_1=\frac{2(+)2\sqrt{3}} {2}=1+\sqrt{3}\\\\x_2=\frac{2(-)2\sqrt{3}} {2}=1-\sqrt{3}$

therefore

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

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

Part 2)

in this problem we have

$x^{2} -6x+4=0$

$a=1\\b=-6\\c=4$

substitute in the formula

$x=\frac{-(-6)(+/-)\sqrt{-6^{2}-4(1)(4)}} {2(1)}$

$x=\frac{6(+/-)\sqrt{20}} {2}$

$x=\frac{6(+/-)2\sqrt{5}} {2}$

$x_1=\frac{6(+)2\sqrt{5}}{2}=3+\sqrt{5}$

$x_2=\frac{6(-)2\sqrt{5}}{2}=3-\sqrt{5}$

therefore

$x^{2} -6x+4=(x-(3+\sqrt{5}))(x-(3-\sqrt{5}))$

$x^{2} -6x+4=(x-3-\sqrt{5})(x-3+\sqrt{5})$

5 0

3 years ago

PLS HELP ME ASAP FOR THIS!! (MUST SHOW WORK!!) + LOTS OF POINTS!!

OlgaM077 [116]

You can see that angle TQV = 90 degrees
Angle RQV = 136 degrees

Angles RQV and UQV are supplementary and equal 180 degrees since they form a straight line. Therefore angle UQV = 180 - 136 = 44 degrees.

TQU + UQV = 90 degrees. Therefore, TQU = 90 - UQV.

90 - 44 = 46 degrees for angle TQU.

8 0

3 years ago

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motikmotik

Answer:

Step-by-step explanation:

DID THE WORK

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4 years ago

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