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3 years ago

6

6th grade math help me please :(

1 answer:

muminat3 years ago

4 0

i think a...... or c :( hope its right

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Can someone please help me with this attachment below

Marizza181 [45]

The first one might be cannot be determined
2- rational
3- irrational

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3 years ago

To rent a certain meeting room, a college charges a reservation fee of $47 and an additional fee of $9.30 per hour. The film clu

dexar [7]

The total number of hours is 5
Why?
equation = 9.30x + 47 = 93.50
9.30x + ( 47 - 47 ) = 93.50 - 47 = 46.5
(9.30x / 9.30) = 46.5 / 9.30 = 5

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4 years ago

HELPPPP PLEASE ASAP ASAP ILL MARK BRAINLIEST :) SHOW YOUR WORK!!!!

Arada [10]

Answer:

4 cm

Step-by-step explanation:

5 0

3 years ago

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

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

so

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

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

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

so

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

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

Product mentally 4 times 5 1/8

Nataly [62]

The answer is 20.5 try it good luck

5 0

3 years ago