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3 years ago

Please help me with this question, image attached.

Mathematics

2 answers:

zaharov [31]3 years ago

7 0

Answer:

<3 = 110

Step-by-step explanation:

The angles are consecutive interior angles. They add up to 180 degrees when the lines are parallel.

<3 + <8 = 180

<3 + 70 =180

Subtract 70 from each side

<3 +70-70=180-70

<3 = 110

Lorico [155]3 years ago

5 0

Answer:

110 degrees.

Step-by-step explanation:

Angles 2 and 3 are supplementary, and angles 8 and 7 are supplementary.

Angles 3 and 7 are equal.

If angle 8 is equal to 70 degrees, angle 3 is equal to 110.

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Which of the following is the scientific notation for the number 0.00001526?

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Answer:

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Step-by-step explanation:

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3 years ago

Use the distributive property of multiplication to find 5×45.<br> Hint: 45=5+40.

alexgriva [62]

Answer:

Step-by-step explanation:

Distributive property: a(b +c) = (a*b) + (a*c)

45 = 40 + 5

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2 years ago

You grab two jelly beans from a jar at the same time. b you draw a card from a deck, replace it, and draw a second. c you draw a

Charra [1.4K]

How is this a question did you re word it

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3 years ago

5(n–2)=5n–10 What is the awnser

Leokris [45]

Answer:

Anything

Step-by-step explanation:

5n - 10 = 5n - 10

It doesn't matter what you put in for n, the equality can never be anything but an equality.

Suppose n = 10,000,000

Then 5*10,000,000 - 10 = 5*10,000,000

6 0

3 years ago

Read 2 more answers

Factor completely. <br> <img src="https://tex.z-dn.net/?f=x%5E%7B8%7D-%5Cfrac%7B1%7D%7B81%7D" id="TexFormula1" title="x^{8}-\fra

Eduardwww [97]

We have 3⁴ = 81, so we can factorize this as a difference of squares twice:

$x^8 - \dfrac1{81} = \left(x^2\right)^4 - \left(\dfrac13\right)^4 \\\\ x^8 - \dfrac1{81} = \left(\left(x^2\right)^2 - \left(\dfrac13\right)^2\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)$

Depending on the precise definition of "completely" in this context, you can go a bit further and factorize $x^2-\frac13$ as yet another difference of squares:

$x^2 - \dfrac13 = x^2 - \left(\dfrac1{\sqrt3}\right)^2 = \left(x-\dfrac1{\sqrt3}\right)\left(x+\dfrac1{\sqrt3}\right)$

And if you're working over the field of complex numbers, you can go even further. For instance,

$x^4 + \dfrac19 = \left(x^2\right)^2 - \left(i\dfrac13\right)^2 = \left(x^2 - i\dfrac13\right) \left(x^2 + i\dfrac13\right)$

But I think you'd be fine stopping at the first result,

$x^8 - \dfrac1{81} = \boxed{\left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)}$

6 0

3 years ago