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Firlakuza [10]
3 years ago
11

Please look at all attachments to see all the graphs. thank you.

Mathematics
1 answer:
ale4655 [162]3 years ago
8 0
Your answer is the first one,k.
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Solve for x please<br> A)-8<br> B)-9<br> C)8<br> D9
netineya [11]

Answer:

My answer is -9.becasiug

7 0
2 years ago
Help please with my "science" hw everything relates to math ugh
sergeinik [125]
1. 3.0 mL * (1 L )/((1000 mL) = 0.003 L

2. 0.24 dag * (10 g)/(1 dag) * (1000 mg)/(1 g) = 2,400.0 mg

3. 1350.0 mm * (1 m)/(1000 mm) = 1.35 m

4. 20.0 kg * (1000 g)/(1 kg) = 20,000.0 g

5. 0.4 cL * (1 L)/(100 cL) * (1 hL)/(100 L) = 0.00004 hL

6. 310.0 m * (1 km)/(1000 m) = 0.31 km


5 0
4 years ago
Can someone PLEASE explain how to simplify square roots with variables and eexponents in them?? I'd also be thankful if you expl
Verizon [17]
X^a/b is  \sqrt[b]{x^a} . The way I memorise that is x^1/3 is the cubic root of x. Do you get it? In that case, x is raised to a power of 1 and the cubic root is practically has a power of 3.
In your example, 

\sqrt[ \frac{3}{2} ]{16 x^4} is practically square rooting each term then cubing them individually. Remember when square-rooting any index you halve it. I'll elaborate:

\sqrt{x^4} = x^{2}
\sqrt{16} = 4
Then cube each,
4^3 = 64 
and ( x^{2} )^3 = x^{6}

As for the 2nd part: you must use the rules of indices.
x^{a}  *  x^{b} =  x^{a+b}
So breaking the question up:

3 * 3 = 9
x^{ \frac{1}{2} } stays as is since the 2nd term does not contain x
now: 
y^{ \frac{4}{3} }  * y^{1} =  y^{ \frac{4}{3} + 1 }  =  y^{ \frac{4}{3} +  \frac{3}{3} }  =   y^{ \frac{7}{3} }
This makes your final answer look like this:
9 x^{ \frac{1}{2} }  y^{ \frac{7}{3} }

I hope that helped and good luck in your test!
4 0
4 years ago
24, 12,6,...<br> Find the 8th term.
Zinaida [17]

Answer:

Divide 24 24 by 8 8 . a4= ...

Step-by-step explanation:

6 0
3 years ago
How do you do pythagorean theorem when you need to fine the A
fredd [130]
Given that Pythagorean Theorum is
A^2 + B^2 = C^2, you need to do the opposite.
To find for A^2, you need to:
Subtract B^2 from C^2. This creates the equation of:
C^2 - B^2 = A^2. This works because A^2 + B^2 is C^2, so the opposite must work.
Now if you're looking for A and not A^2, you need to subtract B^2 from C^2 and then square root the difference.
I hope this helps!
5 0
4 years ago
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