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SOVA2 [1]
3 years ago
15

How do you solve 6(2/3x-3)>3

Mathematics
1 answer:
VARVARA [1.3K]3 years ago
4 0

Answer:

14>3

Step-by-step explanation:

you would need to turn the 3 in the parenthesis into a fraction which would result in 9/3. 2/3 - 9/3= -7/3      

Multiply the 7/3 by 6 and by what I have always been taught, you only multiply the top of the fraction. This would leave it to be 42/3 which would be known as 14.

So therefore the answer would be 14>3

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If x equals 7, then find the matching value for y.<br> -2x-2y=-2
mihalych1998 [28]
Y = -6.
-2(7)-2y=-2

-14-2y=-2

you need to move -14 to the other side by canceling out -14 with +14, and add +14 to -2, giving you -2y=12. -2/12=-6


7 0
3 years ago
Answer the question pls xx
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Step-by-step explanation:

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Radical 6x^2 multiplied by radical 18x^2
Alina [70]

Does radical mean square root?, because radical can mean square root, cube root, fourth root, etc.


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6 0
3 years ago
An Italian sausage is 9 inches long. The chef needs to cut it into 2/3 inch pieces. How many pieces will he be able to cut?
Pavel [41]
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6 0
3 years ago
Write the general polynomial p(x) if its only zeros are 1 4 and -3 with multiplicites 3 2 and 1 respectively what is the degree
kompoz [17]

Answer:

The 6th degree polynomial is p(x) = (x-1)^3(x-4)^2(x+3)

Step-by-step explanation:

Zeros of a function:

Given a polynomial f(x), this polynomial has roots x_{1}, x_{2}, x_{n} such that it can be written as: a(x - x_{1})*(x - x_{2})*...*(x-x_n), in which a is the leading coefficient.

Zero 1 with multiplicity 3.

So

p(x) = (x-1)^3

Zero 4 with multiplicity 2.

Considering also the zero 1 with multiplicity 3.

p(x) = (x-1)^3(x-4)^2

Zero -3 with multiplicity 1:

Considering the previous zeros:

p(x) = (x-1)^3(x-4)^2(x-(-3)) = (x-1)^3(x-4)^2(x+3)

Degree is the multiplication of the multiplicities of the zeros. So

3*2*1 = 6

The 6th degree polynomial is p(x) = (x-1)^3(x-4)^2(x+3)

8 0
3 years ago
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