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Verdich [7]
3 years ago
7

Please help me is this statement true or false 3(36 - 11) = 45?

Mathematics
1 answer:
viva [34]3 years ago
3 0
Hi there! The answer is FALSE

To find the answer we need to work out the parenthesis.
3(36 - 11) =
3 \times 36 - 3 \times 11  =
108 - 33 =
75

75 does not equal 45 and therefore the statement is false.


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Answer:

d= im not sure/ sorry

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Factor completely the given polynomial
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6 0
2 years ago
Ben is 4 times as old as I Ishaan and is also 6years older than ishaan
777dan777 [17]

Answer:

24

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7 0
3 years ago
Read 2 more answers
What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?
Natasha2012 [34]

Answer:

x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}  are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

2x^2-10x-3

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}

General form of quadratic equation is ax^2+bx+c

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

D=\sqrt{(-10)^2-4(2)(-3)}

\Rightarrow D=\sqrt{100+24}=\sqrt{124}

Now substituting D in  x=\frac{b^2\pm\sqrt{D}}{2a} we get

x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}

x=\frac{100\pm\sqrt{124}}{4}

x=\frac{100\pm2\sqrt{31}}{4}

x=\frac{50\pm\sqrt{31}}{2}

Therefore, x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}



5 0
3 years ago
Read 2 more answers
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