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3 years ago

How many watts is 30 kilowatts in 24hrs

1 answer:

8 0

30,00 watts.
Think of the metric system like this: K H D B D C M
Now. you have kilo at the begining and milli at the end. To transition from left to right you multiply by 10 each time. To transition from right to left, you divide by 10 each time.

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Which subset of real numbers does not contain the number 1?

Leni [432]

Answer:

I think the answer might be c or b.I hope this helps

5 0

3 years ago

13 POINTS- please help me

allsm [11]

Answer:

See explanation

Step-by-step explanation:

16. Two parallel lines are cut by transversal. Angles with measures $(6x+20)^{\circ}$ and $(x+100)^{\circ}$ are alternate exterior angles. By alternate exterior angles, the measures of alternate exterior angles are the same:

$6x+20=x+100\\ \\6x-x=100-20\\ \\5x=80\\ \\x=16$

Then

$(6x+20)^{\circ}=(6\cdot 16+20)^{\circ}=116^{\circ}\\ \\(x+100)^{\circ}=(16+100)^{\circ}=116^{\circ}$

17. Two parallel lines are cut by transversal. Angles with measures $(2x+12)^{\circ}$ and $(3x-22)^{\circ}$ are alternate interior angles. By alternate interior angles, the measures of alternate interior angles are the same:

$2x+12=3x-22\\ \\2x-3x=-22-12\\ \\-x=-34\\ \\x=34$

Then

$(2x+12)^{\circ}=(2\cdot 34+12)^{\circ}=80^{\circ}\\ \\(3x-22)^{\circ}=(3\cdot 34-22)^{\circ}=80^{\circ}$

18. Two parallel lines are cut by transversal. Angles with measures $(6x-7)^{\circ}$ and $(5x+10)^{\circ}$ are alternate exterior angles. By alternate interior angles, the measures of alternate exterior angles are the same:

$6x-7=5x+10\\ \\6x-5x=10+7\\ \\x=17$

Then

$(6x-7)^{\circ}=(6\cdot 17-7)^{\circ}=95^{\circ}\\ \\(5x+10)^{\circ}=(5\cdot 17+10)^{\circ}=95^{\circ}$

19. The diagram shows two complementary angles with measures $2x^{\circ}$ and $56^{\circ}$ . The measures of complementary angles add up to $90^{\circ},$ then

$2x+56=90\\ \\2x=90-56\\ \\2x=34\\ \\x=17$

Hence,

$2x^{\circ}=2\cdot 17^{\circ}=34^{\circ}$

Check:

$34^{\circ}+56^{\circ}=90^{\circ}$

20. Angles $\angle 1$ and $\angle 2$ are vertical angles. By vertical angles theorem, vertical angles are congruent, so

$m\angle 1=m\angle 2\\ \\5x+7=3x+15\\ \\5x-3x=15-7\\ \\2x=8\\ \\x=4$

Hence,

$m\angle 1=(5x+7)^{\circ}=(5\cdot 4+7)^{\circ}=27^{\circ}\\ \\m\angle 2=(3x+15)^{\circ}=(3\cdot 4+15)^{\circ}=27^{\circ}$

21. $\angle 5$ and $\angle 8$ are supplementary. The measures of supplementary angles add up to $180^{\circ},$ so

$m\angle 5+m\angle 8=180^{\circ}\\ \\3x-40+7x-120=180\\ \\10x-160=180\\ \\10x=180+160\\ \\10x=340\\ \\x=34$

Therefore,

$m\angle 5=(3x-40)^{\circ}=(3\cdot 34-40)^{\circ}=62^{\circ}\\ \\m\angle 8=(7x-120)^{\circ}=(7\cdot 34-120)^{\circ}=118^{\circ}\\ \\62^{\circ}+118^{\circ}=180^{\circ}$