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3 years ago

2x^2-3x=0 helppppppppppp with steps pls

Mathematics

1 answer:

trapecia [35]3 years ago

8 0

Answer:

x=0, x=3/2

Step-by-step explanation:

2x^2-3x=0

Factor out an x

x(2x-3) =0

Using the zero product property

x = 0 2x-3 =0

Add 3 to each side

2x-3+3 =0+3

2x =3

divide each side by 2

2x/2 = 3/2

x=3/2

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5. Find the first, fourth, and tenth terms of the arithmetic

Lubov Fominskaja [6]

The first, fourth and tenth terms of the arithmetic sequence is $-6, -\frac{27}{5}$ and $-\frac{21}{5}$

Explanation:

The given rule for the arithmetic sequence is $A(n)=-6+(n-1)(\frac{1}{5} )$

We need to determine the first, fourth and tenth terms of the sequence.

To find the first, fourth and tenth terms, let us substitute $n=1,4,10$ in the general rule for the arithmetic sequence.

To find the first term, substitute $n=1$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(1)=-6+(1-1)(\frac{1}{5} )$

$A(1)=-6+(0)(\frac{1}{5} )$

$A(1)=-6$

Thus, the first term of the arithmetic sequence is -6.

To find the fourth term, substitute $n=4$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(2)=-6+(4-1)(\frac{1}{5} )$

$A(2)=-6+(3)(\frac{1}{5} )$

$A(2)=\frac{-30+3}{5}$

$A(2)=\frac{-27}{5}$

Thus, the fourth term of the arithmetic sequence is $-\frac{27}{5}$

To find the tenth term, substitute $n=10$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(10)=-6+(10-1)(\frac{1}{5} )$

$A(10)=-6+(9)(\frac{1}{5} )$

$A(10)=-6+\frac{9}{5}$

$A(10)=-\frac{21}{5}$

Thus, the tenth term of the arithmetic sequence is $-\frac{21}{5}$

Hence, the first, fourth and tenth terms of the arithmetic sequence is $-6, -\frac{27}{5}$ and $-\frac{21}{5}$

3 0

4 years ago

The graph (shown in red) below shows the solution set of which inequality?

mojhsa [17]

Answer:
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Explanation:
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A, B and D do not work with those values, so the answer is C

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3 years ago

What is -7/9 x -2/3 in simplest form

Artyom0805 [142]

Answer:

the answer is 14/27

Step-by-step explanation:

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2 years ago

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What is a+ a/5+ 2/5a

Galina-37 [17]

Answer:

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