Twice the sum of 13 and x is not greater than 15

Mathematics

1 answer:

Kobotan [32]3 years ago

7 0

Answer:

x = 0

Step-by-step explanation:

This is saying that 2(13x) < 15

If you have 2[13(0)] < 15

It will come to 2(0) < 15

This equals 0 < 15 which is true therefore making the x value = 0

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3 years ago

(9x+2)(4x^2+35x-9)=0

pychu [463]

$(9x+2)*(4x^2+35x-9)=0$

we can make this...

$9x+2=0$

and

$4x^2+35x-9=0$

then...

$9x+2=0$

$\boxed{x=-\frac{2}{9}}$

$4x^2+35x-9=0$

$x=\frac{-b\pm\sqrt{b^2-4*a*c}}{2*a}$

$x=\frac{-35\pm\sqrt{35^2-4*4*(-9)}}{2*4}$

$x=\frac{-35\pm\sqrt{1369}}{8}$

$x=\frac{-35\pm37}{8}$

$x_1=\frac{-35+37}{8}=\frac{1}{4}$

$x_2=\frac{-35-37}{8}=-9$

$\boxed{\boxed{S=\{\frac{1}{4},-\frac{2}{9},-9\}}}$

I'm sure about my answer, don't matter the order...

4 0

3 years ago

Read 2 more answers

Does someone know if it is right? Please, if you don't know, it ok, but don't answer. PLEASE!

Firlakuza [10]

Ya I believe that is the answer. I got the exact same thing.