All categories

4 years ago

Using the quadratic formula to solve 5x=6x squared -3, what are the values of x?

1 answer:

7 0

Answer: 5+sqrt(97)/12, 5-sqrt(97)/12 or we write it as [5+/-sqrt(97)]/(12)

5x=6x^2-3

6x^2-5x-3=0

x=-b+/-sqrt(b^2-4ac)/2a

x=-(-5)+/-sqrt((-5^2-4(6)(-3))/2(6)

x= 5+/-sqrt(97)/12

And decimal form is;

x=1.237, -0.4041

If anyone has any questions please feel free to ask and I’ll reply ASAP. Thanks

You might be interested in

I got the answer it is A I just need you to check it pls thank you

andrey2020 [161]

We have a circle of radius 5 m inside a square of side 10 m. The shaded area is:

$A_{\text{Shaded}}=A_{\text{Square}}-A_{\text{Circle}}$

Now, we calculate the area of the square and the circle:

$\begin{gathered} A_{\text{Square}}=10^2=100m^2 \\ A_{\text{Circle}}=\pi\cdot5^2=25\pi m^2 \end{gathered}$

Finally, the shaded area is:

$A_{\text{Shaded}}=100-25\pi\approx21.5m^2$

3 0

1 year ago

Help!!(thank you so much)

mylen [45]

Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the 10. If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.

Or just subtract

9.884 * 10^6 - 1.009 * 10^7=

− 2.05999999 × 10^5

Rounded up is:

− 2.06× 10^5

(❁´◡`❁)

8 0

4 years ago

inverse cosine is used to find to find a given angle measure when we know the ________ and the ______.

Alexxx [7]

Answer: adjacent and hypotenuse

Step-by-step explanation: Cos theta is adjacent/Hypotenuse

4 0

3 years ago

What is the value of the expression i^0 * i^1 * i^2 * i^3 * i^4? A. 1 B. -1. C. i. D. -i

Nastasia [14]

This would be 1 * i * -1 * -i * 1 = -i * -i * 1 = i^2 = -1

3 0

3 years ago

Read 2 more answers

Nadine and Calvin are simplifying the expression (StartFraction r Superscript negative 5 Baseline s Superscript negative 3 Basel

Nat2105 [25]

Answer:

Calvin's first step is to simplify the expression is to apply the quotient of powers to get (r Superscript negative 13 Baseline s Superscript negative 1 Baseline) Superscript negative 4 is the correct step

That is $(\frac{r^{-5}s^{-3}}{r^8s^{-2}})^{-4}=(r^{-13}s^{-1})^{-4}$ Calvin's step is the correct step.Because this is the correct way to do simplify the rational expression. And also because Nadine made a blender mistake in her operations in step

Step-by-step explanation:

Given that Nadine and Calvin are simplifying the expression (StartFraction r Superscript negative 5 Baseline s Superscript negative 3 Baseline Over r Superscript 8 Baseline s Superscript negative 2 Baseline EndFraction) Superscript negative 4

Their expression can be written as below

$(\frac{r^{-5}s^{-3}}{r^8s^{-2}})^{-4}$

Nadine's first step is to simplify the expression is to raise the numerator and denominator to the power of 4 to get StartFraction r Superscript negative 20 Baseline s Superscript negative 12 Baseline Over r Superscript 32 Baseline s Superscript 8 Baseline EndFraction

That is $\frac{r^{20}s^{12}}{r^{-32}s^8}$

Calvin's first step is to simplify the expression is to apply the quotient of powers to get (r Superscript negative 13 Baseline s Superscript negative 1 Baseline) Superscript negative 4

That is $r^{-13}s^{-1}$

Now simplify the given expression to check whose step is correct:

$(\frac{r^{-5}s^{-3}}{r^8s^{-2}})^{-4}$

$=(r^{-5}s^{-3}r^{-8}s^{2})^{-4}$ ( using the property $\frac{1}{a^m}=a^{-m}$ )

$=(r^{-5-8}s^{-3+2})^{-4}$

$=(r^{-13}s^{-1})^{-4}$

Therefore $(\frac{r^{-5}s^{-3}}{r^8s^{-2}})^{-4}=(r^{-13}s^{-1})^{-4}$

Therefore Calvin's first step is to simplify the expression is to apply the quotient of powers to get (r Superscript negative 13 Baseline s Superscript negative 1 Baseline) Superscript negative 4 is the correct step.

That is $(\frac{r^{-5}s^{-3}}{r^8s^{-2}})^{-4}=(r^{-13}s^{-1})^{-4}$ Calvin's step is the correct step .Because this is the correct way to do simplify the rational expression.And also because Nadine made a blender mistake in her operations in step

6 0

3 years ago

Read 2 more answers