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Svetach [21]
3 years ago
9

Is it possible for two numbers to have a difference of 6 and also sum of 6

Mathematics
1 answer:
Snowcat [4.5K]3 years ago
3 0

Definitely.

But the two numbers have to be 6 and zero.

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Steve drove at a constant rate to the beach for a vacation. In the equation below, t is the time in hours it took Steve to drive
My name is Ann [436]
310/62= hours which equals 5! hope this helped a bit.
5 0
3 years ago
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FInd the solution for 4(x - 5)=2(x - 10) +2x
crimeas [40]
Hello!

You solve this like an algebraic equation

You first distribute the 4 and 2

4x - 20 = 2x - 20 + 2x

Combine like terms

4x - 20 = 4x - 20

subtract 4x from both sides

-20 = 0x - 20

Add 20 to both sides

0 = 0

This means that there are infinite solutions to the equation.

Hope this helps!
3 0
3 years ago
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Combine the like terms to create an equivalent -3х – 6+(-1)​
pishuonlain [190]
You can combine the -6 and -1 to -7. So the expression can be simplified as -3x-7
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3 years ago
Simplify the following Questions. (IMPORTANT NOTE!!! #'s 2 AND 7 MUST BE IN SCIENTIFIC NOTATION!!!!!) 1.) (x)²(2xy³)^5 2.) (4x10
poizon [28]

Answer:

Step-by-step explanation:

Hint :

(a*b)^{m}= a^{m}b^{m}\\\\\\(a^{m})^{n}=a^{mn}\\\\\\a^{m}*a^{n}=a^{m+n}\\

1)x^{2}(2xy^{3})^{5}=x^{2}*2^{5}*x^{5}*(y^{3})^{5}\\\\\\=x^{2}*2^{5}*x^{5}*y^{3*5}\\\\=x^{2}*2^{5}*x^{5}*y^{15}\\\\=2^{5}*x^{2+5}*y^{15}\\\\=2^{5}x^{7}y^{15}

2) (4x10^{8})^{2}=4^{2}*x^{2}*(10^{8})^{2}\\\\=16*x^{2}*10^{8*2}\\\\=1.6*10*x^{2}*10^{16}\\\\=1.6x^{2}*10^{16+1}\\\\=1.6x^{2}10^{17}

3)(-h^{4})^{5}=(-h)^{4*5}=-h^{20}\\\\\4)(3xy^{3})^{2}(xy)^{6}=3^{2}x^{2}(y^{3})^{2}x^{6}y^{6}\\\\=9x^{2}y^{3*2}x^{6}y^{6}\\\\\\=9x^{2}y^{6}x^{6}y^{6}\\\\=9x^{2+6}y^{6+6}\\\\=9x^{8}y{12}\\\\\\

5) (p^{9})^{-2}=p^{9*-2}=p^{-18}\\\\\\6)(5k^{2})^{3}=5^{3}(k^{2})^{3}=625k^{6}\\\\7)(7x10^{5})^{2}=7^{2}x^{2}(10^{5})^{2}\\\\=49x^{2}10^{5*2}\\\\=4.9*10^{1}x^{2}10^{10}\\\\=4.9x^{2}10^{10+1}\\\\=4.9x^{2}10^{11}

8)x^{3}(-x^{3}y)^{2}=x^{3}*(-x^3})^{2}y^{2}\\\\\\=x^{3}*(-1)^{2}*x^{2*3}y^{2}\\\\=x^{3}*1*x^{6}y^{2}=x^{3+6}y^{2}\\\\=x^{9}y^{2}\\\\\\9)w^{5}(w^{2})^{-4}=w^{5}w^{2*-4}\\\\=w^{5}w^{-8}\\\\=w^{5-8}=w^{-3}\\\\=\frac{1}{w^{3}}\\\\\\10) (2x^{5})^{4}=2^{4}x^{5*4}\\\\=2^{4}x^{20}\\\\=16x^{20}

5 0
3 years ago
32+b=54 when b = 9 <br> what is the answer
Strike441 [17]

Step-by-step explanation:

I have two step

step 1

32+b=54

32+9=54

41=54

you divide both side by 41 and the answer is 1 and 1

step two

32+b=54

b=54-32

9=22

you divide both side by 9 and the answer is 2,44

7 0
2 years ago
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