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Brrunno [24]
2 years ago
8

Solve this -5-x=-3x-17

Mathematics
1 answer:
drek231 [11]2 years ago
5 0

Answer:

-6

Step-by-step explanation:

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A large number of randomized experiments were conducted to determine whether taking a particular drug regularly would decrease t
DENIUS [597]

Answer:

Step-by-step explanation:

b

7 0
2 years ago
Read 2 more answers
Find the angle between the vectors u=i-9j and v=8i+5j
Katyanochek1 [597]

Answer: \cos^{-1} \left(\frac{-37}{\sqrt{7298}} \right) \approx 115.665^{\circ}

Step-by-step explanation:

u \cdot v=(1)(8)+(-9)(5)=-37\\\\|u|=\sqrt{1^{2}+(-9)^{2}}=\sqrt{82}\\\\|v|=\sqrt{8^{2}+5^{2}}=\sqrt{89}\\\\\cos \theta=\frac{-37}{\sqrt{82}\sqrt{89}}\\\\\theta=\cos^{-1} \left(\frac{-37}{\sqrt{82}\sqrt{89}} \right)\\\\\theta=\boxed{\cos^{-1} \left(\frac{-37}{\sqrt{7298}} \right) \approx 115.665^{\circ}}

7 0
2 years ago
Find the inverse of {(-2, 4), (-3, -1), (2, 2), (3, 4)}. State whether the inverse is a function.
Flura [38]

Answer:

(-4, 2), (-1, -3), (2,0), (4,3).

8 0
2 years ago
How to do the problem
MrMuchimi
Order of operations are:
<span>1. Parentheses (simplify inside 'em)
2. Exponents
3. Multiplication and Division (from left to right)
<span>4. Addition and Subtraction (from left to right)

So, on your first one:
Parentheses first:
8(3+4)-2*8/(5-3)
8(7)-2*8/(2)

There are no exponents so multiplication and division:
56-16/2
56-8

Finally, addition and subtraction:
56-8=48

Your second problem is a bit more complicated but follows the same rules:
</span></span>( 8^{2}+(13-4) ^{2} )/5
<span><span>
Parentheses first:
</span></span>( 8^{2}+ 9^{2})/5)
<span><span>
Now, exponents. Even though they are inside parentheses, we can't go further until we simply those.
(64+81)/5

Now back to parentheses:
(145)/5

Division:
145/5=29

Hope that helps</span></span>
8 0
3 years ago
Which is best estimate of 90/7 ÷ 1 3/4?<br><br> 2<br><br> 6<br><br> 12<br><br> 24
Vsevolod [243]

Answer:

6

Step-by-step explanation:

90/7 divided by 1 3/4 is 7.34 so the closest one to it is 6

8 0
3 years ago
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