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3 years ago

26.013 powers of ten

Mathematics

2 answers:

White raven [17]3 years ago

3 0

Step-by-step explanation: So ig this wat you ASKED FOR?

natulia [17]3 years ago

3 0

Hope this image helps

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Find the midpoint given the endpoints (7,-9)<br> and (-10,8).

DanielleElmas [232]

Answer:

(-3/2, -1/2)

Step-by-step explanation:

Midpoint formula: (x1 + x2 / 2), (y1 + y2 / 2)

(7 + (-10) / 2), (-9 + 8 / 2)

(-3/2, -1/2)

3 0

3 years ago

Simplify the expression 21b^4 - 3b by combining like terms

IRISSAK [1]

<u>Answer</u>:

❁ Hello! ❁

21b⁴ - 3b = 3b(7b³ - 1)

6 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

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

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

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

4 years ago

CAN SOMEBODY PLEASE HELP!!! I will cashapp you

mr_godi [17]

Answer:

its wrong

Step-by-step explanation:

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3 years ago

How would you write out 3 x 6 + 7 x 4 = 46?

aleksandrvk [35]

Answer:

dceoi hchis hurhgj

Step-by-step explanatiicdeon:

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3 years ago