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I am Lyosha [343]
2 years ago
8

The element silicon has a molar mass of 28.09 g/mol. Calculate the mass of 0.15 mol of Silicon.

Mathematics
1 answer:
Sav [38]2 years ago
3 0

The answer is 28.09 g. I hope this helped.

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Need help with number 7
padilas [110]

Answer:

hi there!

the correct answer to this question is: 24 crackers

Step-by-step explanation:

you add (2/5) and (1/5) together to get (3/5)

then you multiply 3/5 by 40 to get 24

6 0
3 years ago
Could someone confirm if I was correct or not?
Andrei [34K]
Correct! Excellent Job
7 0
3 years ago
Read 2 more answers
What is the value of x?
soldi70 [24.7K]

Answer:

5\sqrt{3}

Step-by-step explanation:

A triangle with angles of 30-60-90 will always have sides that measure in the following proportions:

The side opposite the 30* angle = a

The side opposite the 60* angle = a\sqrt{3}

The side opposite the 90* angle = 2a

(I used "a" here because your question uses "x" and I don't want to confuse things, but usually people use "x" rather than "a")

They gave us that the hypotenuse (the side opposite the 90* angle) is 10, from there we can figure out the other sides:

Since the side opposite 90* is 2a, and 2a = 10, divide 10 by 2 and you get a = 5.

Once we know what "a" is, we can fill in the rest:

30* = 5

60* = 5\sqrt{3}

90* = 10

3 0
3 years ago
Please help me.. im stuck​
nalin [4]

Answer: x = 2/3 and y = 2/3

Step-by-step explanation:

I've attached the solution.

From the solution, you get y = -5x + 4 and

y = x.

Now,

Put y = x in y = -5x + 4.

x = -5x + 4

6x = 4

x = 4/6

x = 2/3

So, y = 2/3 because y = x.

6 0
3 years ago
If JK=20-x^2, KL=2-x, and JL=10 find x
kherson [118]
|JK|=20-x^2;\ |KL|=2-x;\ |JL|=10\\D:20-x^2 > 0\ and\ 2-x > 0\ and\ 20-x^2 < 10\ and\ 2-x < 10\\x\in(-\sqrt{20};\ \sqrt{20})\ and\  x < 2\ and\\x\in(-\infty; -\sqrt{10})\ \cup\ (\sqrt{10};\ \infty)\ and\ x >-8\\\\x\in(-\sqrt{20};-\sqrt{10})\\\\|JL|=|JK|+|KL|\\\\20-x^2+2-x=10\\\\-x^2-x+22=10\ \ \ \ \ |subrtact\ 10\ from\ both\ sides\\\\-x^2-x+12=0\\\\-(x^2+x-12)=0\\\\x^2+x-12=0\\\\x^2+4x-3x-12=0\\\\x(x+4)-3(x+4)=0\\\\(x+4)(x-3)=0\iff x+4=0\ or\ x-3=0\\\\x=-4\in D\ or\ x=3\notin D\\\\Answer:\boxed{x=-4}
6 0
3 years ago
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