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EastWind [94]
3 years ago
8

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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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

3 0
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
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