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3 years ago

What is the answer ?

Mathematics

1 answer:

andrezito [222]3 years ago

5 0

Answer:your answer is d

Step-by-step explanation:

Plz mark brainliest

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3 years ago

Which strategy can be used to multiply 6 x 198 mentally?

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3 years ago

If you vertical stretch the quadratic parent function F(x)=x^2By multiplying by seven is what is the equation of the new functio

myrzilka [38]

Answer:

g(x) = 7x^2

Step-by-step explanation:

Given a function f(x), the function kf(x) is stretched by a factor of k. In this case, if we stretch the function f(x) = x^2 by a factor of seven, the new function is going to be the base function multiplied by 7, as follows:

g(x) = 7x^2

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3 years ago

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

compare the right-hand and left-hand derivatives to show that the function is not differentiable at the point P. find all points

Softa [21]

$f(x)= \left \{ {{ \sqrt{x} } \atop {2x-1}} \right. 
\\
\\ f'(x)= \left \{ {{ \frac{1}{2\sqrt{x}} } \atop {2}} \right. 
\\
\\ f'(1)= \left \{ {{ \frac{1}{2}} \atop {2}} \right. 
\\
\\ f'(1^-) \neq f'(1^+)$

Therefore, the function is not differentiable at x = 1.

$f(x)= \sqrt{x} 
\\
\\f'(x)= \frac{1}{2 \sqrt{x} } 
\\
\\f'(0)=\frac{1}{2 \sqrt{0} } = \infty$

Therefore, the function is not differentiable at x = 0.

3 0

3 years ago