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wolverine [178]
4 years ago
8

How do you find the least common multiple of these two expressions. 10x^5v^8 and 25x^4v^7u^2

Mathematics
2 answers:
RoseWind [281]4 years ago
8 0
10x^5v^8=5x^4v^7\cdot2xv\\\\25x^4v^7u^2=5x^4v^7\cdot5u^2\\\\LCM(10x^5v^8;\ 25x^4v^7u^2)=5x^4v^7\cdot2xv\cdot5u^2=50x^5v^8u^2
Ahat [919]4 years ago
4 0
10x^{5} v^{8}= 1*2*5*x*x*x*x*x*v*v*v*v*v*v*v*v \\ 25 x^{4} v^{7} u^{2}=1*5*5*x*x*x*x*v*v*v*v*v*v*v*u*u \\
Underline all the numbers which are common in both the factorisations. Write them down:
5 x^{4} v^{7}
There, you have your LCM of both the expressions.
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A fish tank is three-fourths full.
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Step-by-step explanation:

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2 years ago
What is the equation for a line that has a slope of 3 and passes through (-1, 9)
krok68 [10]

Answer:

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7 0
3 years ago
Solve the system of equations using substitution or elimination. Show all work for credit.
Tomtit [17]
Using substitution:
first you have to express one variable in terms of the other, in this we can express y in terms of x:
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Since both expressions are equal to y, you have to equal both expressions like this:
\frac{  - \frac{ 1}{2}x + 2 }{3}  =  - x + 9
Now you can solve the equation:
-  \frac{1}{2} x + 2 = 3( - x + 9) \\   - \frac{ 1}{2}x  + 2 =  - 3x + 27 \\  \\  -  \frac{1}{2} x + 3x = 27 - 2 \\   \frac{5}{2} x = 25 \\ x =  \frac{(25)(2)}{5}  \\ x = 10
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4 0
3 years ago
Look at the Schoology problem provided below.
lys-0071 [83]

Answer:

k = 4 units

Step-by-step explanation:

Using Pythagoras' identity on the right triangle.

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is

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k² = 16 ( take the square root of both sides )

k = \sqrt{16} = 4

6 0
3 years ago
Read 2 more answers
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