All categories

3 years ago

How do you solve |5x-4|>6

Mathematics

1 answer:

belka [17]3 years ago

5 0

Answer:

x < - $\frac{2}{5}$ or x > 2

Step-by-step explanation:

inequalities of the form | x | > a always have solutions of the form

x < - a or x > a, hence

5x - 4 < - 6 or 5x - 4 > 6 ( add 4 to both sides in both inequalities )

5x < - 2 or 5x > 10 ( divide both sides by 5 in both inequalities )

x < - $\frac{2}{5}$ or x > 2

The solutions can be expressed in interval notation as

x ∈ (- ∞, - $\frac{2}{5}$ ) ∪ (2, ∞ )

You might be interested in

Plzzz help ASAP image above

Elanso [62]

Answer:

Step-by-step explanation:

When you translate a point down you're just moving 2 units down. When it's reflected across the y or x axis, it'll be like a mirror.

3 0

4 years ago

The curve with equation y=e^2x/4+3e^x has one stationary point.find the exact value of the coordinates of this point

Softa [21]

$y=\dfrac{e^{2x}}{4+3e^x}$
$\implies y'=\dfrac{2e^{2x}(4+3e^x)-3e^xe^{2x}}{(4+3e^x)^2}$
$\implies y'=\dfrac{8e^{2x}+3e^{3x}}{(4+3e^x)^2}$

Stationary points occur where the derivative is zero. The denominator is positive for all $x$ , so we only need to worry about the numerator.

$8e^{2x}+3e^{3x}=e^{2x}(8+3e^x)=0$

$e^{2x}>0$ for all $x$ , so we can divide through:

$8+3e^x=0\implies e^x=-\dfrac83$

But $e^x>0$ for all $x\in\mathbb R$ , so this function has no stationary points...

I suspect there may be a typo in the question.

6 0

3 years ago

Whats the intrgral of <img src="https://tex.z-dn.net/?f=%20%5Cint%20%20%5Cfrac%7Bx%5E2%2Bx-3%7D%7B%28x%5E3%2Bx%5E2-4x-4%29%5E2%7

rosijanka [135]

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$

Notice that $x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1)$ . Decompose the integrand into partial fractions:

$\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}$
$=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}$

Integrating term-by-term, you get

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$
$=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C$

5 0

4 years ago

Solve step by step -18=2(-q+4)

Mila [183]

Answer:

q=13

Step-by-step explanation:

First, you can multiply to distribute 2 over -9 and 4.

-18=2(-q+4)

-18=-2q+8

Then, you can subtract 8 from both sides.

-18=-2q+8

-26=-2q

Finally, you can divide -2 on both sides.

13=q

Now, you can check by plugging in 13 for the value of q in the original equation.

-18=2(-(13)+4)

-18=2(-13+4)

-18=2(-9)

-18=-18

Hope this helps you! :)

6 0

3 years ago

Solve -2 (t- 1) = 18

Alla [95]

Answer:

t=-8

Step-by-step explanation:

-2 (t- 1) = 18

Divide each side by -2

-2 (t- 1)/-2 = 18/-2

t-1 = -9

Add 1 to each side

t-1+1 = -9+1

t = -8

5 0

3 years ago

Read 2 more answers