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Alika [10]
4 years ago
5

What are the zeros of the function represents by the quadratic expression 2x^2+x-3?

Mathematics
1 answer:
larisa [96]4 years ago
3 0

Answer:

\large\boxed{x=-1.5\ or\ x=1}

Step-by-step explanation:

The zeros of f(x) = 2x² + x - 3:

2x^2+x-3=0\\2x^2+3x-2x-3=0\\x(2x+3)-1(2x+3)=0\\(2x+3)(x-1)=0\iff2x+3=0\ \vee\ x-1=0\\\\2x+3=0\qquad\text{subtract 3 from both sides}\\2x=-3\qquad\text{divide both sides by 2}\\x=-1.5\\\\x-1=0\qquad\text{add 1 to both sides}\\x=1

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5 0
3 years ago
Read 2 more answers
Plzz help Solve for x x ÷3 3/10 =2 2/5
Svet_ta [14]

Answer:

\huge\boxed{x=7\dfrac{23}{25}}

Step-by-step explanation:

x\div3\dfrac{3}{10}=2\dfrac{2}{5}\\\\\text{convert the mixed number to the impropper fraction}\\\\3\dfrac{3}{10}=\dfrac{3\cdot10+3}{10}=\dfrac{33}{10}\\\\2\dfrac{2}{5}=\dfrac{2\cdot5+2}{5}=\dfrac{12}{5}\\\\x\div\dfrac{33}{10}=\dfrac{12}{5}\\\\x\times\dfrac{10}{33}=\dfrac{12}{5}\qquad\text{multiply both sides by}\ \dfrac{33}{10}\\\\x\times\dfrac{10\!\!\!\!\!\diagup}{33\!\!\!\!\!\diagup}\times\dfrac{33\!\!\!\!\!\diagup}{10\!\!\!\!\!\diagup}=\dfrac{12}{5}\times\dfrac{33}{10}\\\\x=\dfrac{396}{50}

x=\dfrac{198}{25}\\\\x=\dfrac{175+23}{25}\\\\x=\dfrac{175}{25}+\dfrac{23}{25}\\\\x=7\dfrac{23}{25}

3 0
3 years ago
Please answer this correctly
Ymorist [56]

B; 25.65+13.23+6.35=45.23

To do this mentally just add the 2 decimals places together first (1.23). Then just add the whole numbers (44). Then add the 2 numbers together, 44+1.23=45.23

8 0
3 years ago
Compute a value for t that satisfies the equation below. "-2/3" t - 2 = - 3
Dima020 [189]

Answer:

t = 3/2

Step-by-step explanation:

Instead of randomly guessing values of "t" that will satisfy the equation, you can easily find the correct value by solving the equation in terms of "t". In other words, you can set the equation equal to "t" to find the final answer.

(-2/3)t - 2 = -3                             <----- Original equation

(-2/3)t = -1                                   <----- Add 2 to both sides

t = 3/2                                        <----- Divide both sides by -2/3

You can check this value by plugging it into "t" and determining whether both sides of the equations will be equal.

(-2/3)t - 2 = -3                            <----- Original equation

(-2/3)(3/2) - 2 = -3                      <----- Plug 3/2 into "t"

-6/6 - 2 = -3                               <----- Multiply -2/3 and 3/2

-1 - 2 = -3                                    <----- Simplify -6/6

-3 = -3                                        <----- Subtract

6 0
2 years ago
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