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2 years ago

14. If f(x) = 3x - 4 and g(x) = x2 - 4x + 6, then f(x) - g(x) for what values of x?

Mathematics

1 answer:

Ugo [173]2 years ago

6 0

Answer:

X = 2

Step-by-step explanation:

Read more on parallelogram;

brainly.com/question/970600

#SPJ1

6 0

1 year ago

Using the diagram on the right, find the length of GT and TA.

Art [367]

Answer:

$GT=10\ units$ and $TA=6\ units$

Step-by-step explanation:

we know that

In the diagram

Triangles GRT and GEA are similar by AA Similarity Postulate

Remember that

If two figures are similar, then the ratio of its corresponding sides is proportional

$\frac{GR}{GE}=\frac{GT}{GA}$

we have

$GR=5\ units$

$GE=5+3=8\ units$

$GT=16-x\ units$

$GA=16\ units$

substitute

$\frac{5}{8}=\frac{16-x}{16}\\ \\5*16=8*(16-x)\\ \\80=128-8x\\ \\8x=128-80\\ \\x=48/8\\ \\x=6\ units$

therefore

$GT=16-x=16-6=10\ units$

$TA=x=6\ units$

5 0

3 years ago

What is f(-2) for f(x)=1/2x^2

GarryVolchara [31]

Answer: 2

Steps:
f(-2) = 1/2x^2
f(-2) = 1/2(-2)^2
f(-2) = 1/2(4)
f(-2) = 2

7 0

3 years ago

kotykmax [81]

Answer:

16.5 hours

Step-by-step explanation:

The bicycle costs $240. Eva already has $42. She is making $12 per hour.

240 = 12x + 42

Solve for x to find out how many hours Eva needs to work to buy the bicycle.

240 = 12x + 42

240 - 42 = (12x + 42) - 42

198 = 12x

198/12 = (12x)/12

16.5 = x

Eva needs to work 16.5 hours to be able to buy the bikw.

6 0

3 years ago

adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the

Zigmanuir [339]

Answer:

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$ and $\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

Step-by-step explanation:

The equation of the isotope decay is:

$\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }$

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

$\tau = \frac{5568\,years}{\ln 2}$

$\tau \approx 8032.926\,years$

The decay time is:

$t = 1315\,years + 2007\,years \pm 13\,years$ (There is no a year 0 in chronology).

$t = 3335 \pm 13\,years$

Lastly, the relative amount is estimated by direct substitution:

$\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$

$\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

4 0

2 years ago