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Anarel [89]
1 year ago
12

What is the slope of LaTeX: f\left(x\right)=3x-1

Mathematics
1 answer:
arlik [135]1 year ago
7 0

To find the slope of the expression:

f(x)=y=3x-1

We need to remember that this is the Slope-intercept Form of the line equation:

y=mx+b

Where

m = slope

b is the y-intercept.

Therefore, the slope of the line equation above is m = 3.

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Which inequality is true?
vovangra [49]

Answer:

D |-11 | >-25

Step-by-step explanation:

|-11 | = 11

11 > -25

3 0
2 years ago
PLSSS HELP
Dennis_Churaev [7]

The correct standard form of the equation of the parabola is:

(x+3)^{2}= 4(y - 3).

<h3 /><h3>What is a parabola?</h3>

An equation of a curve that has a point on it that is equally spaced from a fixed point and a fixed line is referred to as a parabola. The parabola's fixed line and fixed point are together referred to as the directrix and focus, respectively. It's also crucial to remember that the fixed point is not located on the fixed line. A parabola is a locus of any point that is equally distant from a given point (focus) and a certain line (directrix). An essential curve of the coordinate geometry's conic sections is the parabola.

For the given question,

Vertex of parabola is (-3,3)

Thus, the equation of the parabola is:

(x-h)^{2}=4(y-k)

(x+3)^{2}= 4(y-3)

Learn more about parabolas here:

brainly.com/question/64712

#SPJ1

4 0
2 years ago
What is the measure of ∠F? Irregular quadriateral. Reading clockwise, the angles are labeled one hundred thirty-two degrees, six
Delvig [45]
Total of all angles should equal 360 degrees.  So 360-132-68-66=94, angle F is 94 degrees.
5 0
3 years ago
Find the real solution(s) of the equation. Round your answer to two decimal places when appropriate. 5x^6=30
Elina [12.6K]

Answer:

The real solutions are

x=\sqrt[6]{6}\approx 1.35\\\\\:x=-\sqrt[6]{6}\approx -1.35

Step-by-step explanation:

The solution, or root, of an equation is any value or set of values that can be substituted into the equation to make it a true statement.

To find the real solutions of the equation 5x^6=30:

\mathrm{Divide\:both\:sides\:by\:}5\\\\\frac{5x^6}{5}=\frac{30}{5}\\\\\mathrm{Simplify}\\\\x^6=6\\\\\mathrm{For\:}x^n=f\left(a\right)\mathrm{,\:n\:is\:even,\:the\:solutions\:are\:}x=\sqrt[n]{f\left(a\right)},\:-\sqrt[n]{f\left(a\right)}\\\\x=\sqrt[6]{6}\approx 1.35\\\\\:x=-\sqrt[6]{6}\approx -1.35

7 0
3 years ago
the half life of c14 is 5730 years. Suppose that wood found at an archeological excavation site contains about 35% as much C14 a
Furkat [3]

Answer:

The wood was cut approximately 8679 years ago.

Step-by-step explanation:

At first we assume that examination occured in 2020. The decay of radioactive isotopes are represented by the following ordinary differential equation:

\frac{dm}{dt} = -\frac{m}{\tau} (Eq. 1)

Where:

\frac{dm}{dt} - First derivative of mass in time, measured in miligrams per year.

\tau - Time constant, measured in years.

m - Mass of the radioactive isotope, measured in miligrams.

Now we obtain the solution of this differential equation:

\int {\frac{dm}{m} } = -\frac{1}{\tau}\int dt

\ln m = -\frac{1}{\tau} + C

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} } (Eq. 2)

Where:

m_{o} - Initial mass of isotope, measured in miligrams.

t - Time, measured in years.

And time is cleared within the equation:

t = -\tau \cdot \ln \left[\frac{m(t)}{m_{o}} \right]

Then, time constant can be found as a function of half-life:

\tau = \frac{t_{1/2}}{\ln 2} (Eq. 3)

If we know that t_{1/2} = 5730\,yr and \frac{m(t)}{m_{o}} = 0.35, then:

\tau = \frac{5730\,yr}{\ln 2}

\tau \approx 8266.643\,yr

t = -(8266.643\,yr)\cdot \ln 0.35

t \approx 8678.505\,yr

The wood was cut approximately 8679 years ago.

5 0
3 years ago
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