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3 years ago

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vinnie hacker shouldn't have followers at all and he doesn't appreciate what his followers do for him if it weren't for us he wo

uld be dating a trump supporter

2 answers:

natulia [17]3 years ago

7 0

shut the fawk up... no one asked

ipn [44]3 years ago

3 0

Answer:

...

Step-by-step explanation:

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Given: g(x) = 4 - (x + 3)^2<br>find the x-intercept

Nata [24]

X intercept is when the function (y) = 0.

0 = 4 - (x+3)^2

(x+3)(x+3) = 4

x^2 + 6x + 9 = 4

x^2 + 6x + 5 = 0

To factor, we find what multiplies to 5 and adds to 6. In this case, it’s 5 and 1.

(x+5)(x+1) = 0

x values of the intercepts are -5 and -1.
The coordinates are (-5,0) and (-1,0).

4 0

3 years ago

Read 2 more answers

Select ALL the correct answers.

Orlov [11]

Answer:

r(n)=3n-1

Step-by-step explanation:

4 0

4 years ago

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Using the point-slope formula, what is the equation of the line that passes through the points (1,5) and (0, 0)?

DerKrebs [107]

Answer:

y =5x

Step-by-step explanation:

$y_{1}-y_{2} = m(x_{1} - x_{2} )$

(5 - 0) = m(1 - 0)

5 = m(1)

5 = m

$y_{1}-y_{2} = m(x_{1} - x_{2} )$

y - 5 = 5(x - 1)

y - 5 = 5x - 5

y = 5x -5 + 5

y = 5x

7 0

3 years ago

Find the limit of the function by using direct substitution.

serg [7]

Answer:

Option a.

$\lim_{x \to \frac{\pi}{2}}(3e)^{xcosx}=1$

Step-by-step explanation:

You have the following limit:

$\lim_{x \to \frac{\pi}{2}{(3e)^{xcosx}$

The method of direct substitution consists of substituting the value of $\frac{\pi}{2}$ in the function and simplifying the expression obtained.

We then use this method to solve the limit by doing $x=\frac{\pi}{2}$

Therefore:

$\lim_{x \to \frac{\pi}{2}}{(3e)^{xcosx} = \lim_{x\to \frac{\pi}{2}}{(3e)^{\frac{\pi}{2}cos(\frac{\pi}{2})}$

$cos(\frac{\pi}{2})=0\\$

By definition, any number raised to exponent 0 is equal to 1

So

$\lim_{x\to \frac{\pi}{2}}{(3e)^{\frac{\pi}{2}cos(\frac{\pi}{2})} = \lim_{x\to \frac{\pi}{2}}{(3e)^{\frac{\pi}{2}(0)}\\\\$

$\lim_{x\to \frac{\pi}{2}}{(3e)^{0}} = 1$

Finally

$\lim_{x \to \frac{\pi}{2}}(3e)^{xcosx}=1$

6 0

3 years ago

Verify that the roots of 5x²- 6x -2 = 0 are <img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%20%2B%20%5Csqrt%7B19%7D%20%7D%7B5%7D%2

Mice21 [21]

Answer:

Proof below.

Step-by-step explanation:

<u>Quadratic Formula</u>

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

<u>Given quadratic equation</u>:

$5x^2-6x-2=0$

<u>Define the variables</u>:

a = 5
b = -6
c = -2

<u>Substitute</u> the defined variables into the quadratic formula and <u>solve for x</u>:

$\implies x=\dfrac{-(-6) \pm \sqrt{(-6)^2-4(5)(-2)}}{2(5)}$

$\implies x=\dfrac{6 \pm \sqrt{36+40}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{76}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{4 \cdot 19}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{4}\sqrt{19}}{10}$

$\implies x=\dfrac{6 \pm2\sqrt{19}}{10}$

$\implies x=\dfrac{3 \pm \sqrt{19}}{5}$

Therefore, the exact solutions to the given <u>quadratic equation</u> are:

$x=\dfrac{3 + \sqrt{19}}{5} \:\textsf{ and }\:x=\dfrac{3 - \sqrt{19}}{5}$

Learn more about the quadratic formula here:

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brainly.com/question/27953354

3 0

2 years ago

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