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kari74 [83]
3 years ago
15

How to solve 6x^2+5x-4=0

Mathematics
1 answer:
Vesna [10]3 years ago
6 0
6x^2+5x-4=0\\\\6x^2+8x-3x-4\\\\2x(3x+4)-1(3x+4)=0\\\\(3x+4)(2x-1)=0\iff3x+4=0\ or\ 2x-1=0\\\\3x=-4\ or\ 2x=1\\\\\boxed{x=-\frac{4}{3}\ or\ x=\frac{1}{2}}

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In the triangle below,<br> y = [ ? ] cm. Round to the<br> nearest tenth.
sattari [20]

Answer:

The answer is

<h2>12.3 cm</h2>

Step-by-step explanation:

Since the triangle is a right angled triangle we can use trigonometric ratios to find y

To find y we use cosine

cos∅ = adjacent / hypotenuse

From the question

y is the adjacent

The hypotenuse is 15

So we have

\cos(35)  =  \frac{y}{15}  \\ y = 15 \cos( 35 )  \\ y = 12.28728

We have the final answer as

<h3>12.3 cm to the nearest tenth</h3>

Hope this helps you

7 0
3 years ago
What is 571,951 in written form
Natali [406]
Five hundered seventy-one thousand nine hundred and fifty one
5 0
3 years ago
Help please This is 8th grade made. I put 13 but Idk if it's correct
Rom4ik [11]
The answer is the square root of (12^2 + 5^2), which is 13. You're correct
5 0
3 years ago
For <img src="https://tex.z-dn.net/?f=e%5E%7B-x%5E2%2F2%7D" id="TexFormula1" title="e^{-x^2/2}" alt="e^{-x^2/2}" align="absmiddl
nevsk [136]
I'm assuming you're talking about the indefinite integral

\displaystyle\int e^{-x^2/2}\,\mathrm dx

and that your question is whether the substitution u=\dfrac x{\sqrt2} would work. Well, let's check it out:

u=\dfrac x{\sqrt2}\implies\mathrm du=\dfrac{\mathrm dx}{\sqrt2}
\implies\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt2\int e^{-(\sqrt2\,u)^2/2}\,\mathrm du
=\displaystyle\sqrt2\int e^{-u^2}\,\mathrm du

which essentially brings us to back to where we started. (The substitution only served to remove the scale factor in the exponent.)

What if we tried u=\sqrt t next? Then \mathrm du=\dfrac{\mathrm dt}{2\sqrt t}, giving

=\displaystyle\frac1{\sqrt2}\int \frac{e^{-(\sqrt t)^2}}{\sqrt t}\,\mathrm dt=\frac1{\sqrt2}\int\frac{e^{-t}}{\sqrt t}\,\mathrm dt

Next you may be tempted to try to integrate this by parts, but that will get you nowhere.

So how to deal with this integral? The answer lies in what's called the "error function" defined as

\mathrm{erf}(x)=\displaystyle\frac2{\sqrt\pi}\int_0^xe^{-t^2}\,\mathrm dt

By the fundamental theorem of calculus, taking the derivative of both sides yields

\dfrac{\mathrm d}{\mathrm dx}\mathrm{erf}(x)=\dfrac2{\sqrt\pi}e^{-x^2}

and so the antiderivative would be

\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt{\frac\pi2}\mathrm{erf}\left(\frac x{\sqrt2}\right)

The takeaway here is that a new function (i.e. not some combination of simpler functions like regular exponential, logarithmic, periodic, or polynomial functions) is needed to capture the antiderivative.
3 0
3 years ago
Simply square roots​
Nadya [2.5K]

Answer:

1. -16 Radical(5i)

2. -10 Radical(5i)

3. -8 Radical(5i)

4. -4 Radical(5i)

Hope this helps :)

8 0
2 years ago
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