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3 years ago

Please help me answer a,b,c as I’m super confused and I’m reslky panicking.sos

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

7 0

Answer:

A) 2:1

B) 2:5

C) 3:8

D) 5:2

E) 7:3

F) 7:1

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What is the answer and how do I do it

telo118 [61]

Answer:

I can't really see the problem but i think its y=6

Step-by-step explanation:

If its y=6, your line is just going to be a horizontal line that goes through the 6 on the y axis

the opposite is true if you had x=6. then it would be a vertical line going through the 6 on the x axis.

3 0

4 years ago

Still confusing to me!!!!

juin [17]

Answer:

(a) 2(x -a)

(b) 4x +2h

Step-by-step explanation:

Fill in the function arguments and simplify.

$\dfrac{f(x)-f(a)}{x-a}=\dfrac{(2x^2-7)-(2a^2-7)}{x-a}\\\\=\dfrac{2x^2-7-2a^2+7}{x-a}=\dfrac{2(x^2-a^2)}{x-a}\\\\=\dfrac{2(x-a)(x+a)}{(x-a)}=2(x+a)$

$\dfrac{f(x+h)-f(x)}{h}=\dfrac{(2(x+h)^2-7)-(2x^2-7)}{h}\\\\=\dfrac{2x^2+4xh+2h^2-7-2x^2+7}{h}=\dfrac{4xh+2h^2}{h}\\\\=4x+2h$

8 0

4 years ago

Plz help me with this I beg you

zimovet [89]

The answer I’m pretty sure will be -4. (-4+12-36)/7 = -28/7.

4 0

3 years ago

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Which of the following has a graph that is a straight line? (4 points)

nlexa [21]

The equation 2 has a graph which is a straight line.

Why?

We can know which of the given equations has a graph which is a straight line just checking the exponents of the variables.

We must remember that every variable that has an exponent equal or higher than 2 (quadratic) will not have a straight line as a graphic.

So, checking the exponents from the given equations, we have:

$Equation_{1} =x^{3}=NotLinear\\Equation_{2}=x=Linear\\Equation_{3}=y^{2}=Quadratic\\Equation_{4}=x^{2}=Quadratic$

Hence, we can see that the only equation that has a linear term (straight line graph), is the second equation.

Have a nice day!

Note: I have attached a image for better understanding.

8 0

3 years ago

The function f(x) = x^2- 4x + 3 has a relative minimum located at which of the following x values?

Luda [366]

Given a quadratic equation $y=ax^2+bx+c$ , the x coordinate of the minumum/maximum is given by

$x=-\dfrac{b}{2a}$

So, in your case, the x coordinate of the minimum is

$x=-\dfrac{-4}{2}=2$

4 0

3 years ago

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