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2 years ago

Whats the vertex of the graph y=-x^2

1 answer:

6 0

(0,0) you could use x=-b/2a but since your b term is nonexistant the whole term goes to 0 for the x value. Plug x=0 into the original equation which gives you another 0. ezpz

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Solve each absolute value inequality. Graph the solutions on a number line. |x-7|>1

N76 [4]

Answer:

x>8 and x<6

Step-by-step explanation:

well first the values of x-7 have to be greater than 1 or, less than -1

because if x-7=-2 then |-2|=2>1 satisfying the equation

so we have

$x-7>1\\\\x-7+7>1+7\\\\x>8$

and we have

$x-7$

to graph it it goes from (-∞,6)∪(8,∞)

8 0

3 years ago

What is:<br> <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B4x%5E%7B2%7D%20-20x%2B25%20" id="TexFormula1" title=" \sqrt{4x^{2} -

Firdavs [7]

We are given :

$\sqrt{4x^{2}-20x+25}$

Step 1: factor the part inside square root

The function given inside square root is of quadratic form.

So let us try to factorise it using AC method.

Here A*C = 4*25 = 100

so we have to find factors of 100 that add up to give -20.

the two factors are -10 and -10.

Rewriting the function :

$4x^{2} -20x+25$

= $4x^{2} -10x-10x+25$

= $2x(2x-5) - 5(2x-5)$

= $(2x-5)^{2}$

Step 2:

Now we take square root of the factorised form

$\sqrt{(2x-5)^2}$

= $2x-5$

Answer : (2x-5)

3 0

3 years ago

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Temka [501]

Answer:

It diverges.

Step-by-step explanation:

We are given the inetegral: $\int\limits^{\infty}_2 \frac{1}{x} (\ln x)^2 dx$

$\int\limits^{\infty}_2 \frac{1}{x} (\ln x)^2 dx=\int\limits^{\infty}_2 (\ln x)^2 d(\ln x)=\\\\=\lim_{t \to \infty} \int\limits^t_2 (\ln x)^2d(\ln x)=\lim_{t \to \infty} \frac{(\ln t)^3}{3} |^t_2=\infty-\frac{(\ln 2)^3}{3} =\infty$