All categories

2 years ago

HELP ASAP WHICH FUNCTION BELOW HAS THE GREATER RATE OF CHANGE THX AND PLZ

Mathematics

1 answer:

Sophie [7]2 years ago

4 0

Answer:

Step-by-step explanation:

B has a slope of 7 while A has a slope of 6.

slope of B 14-7/2-1 = 7

A slope is y= mx+b and m is the slope. so 6.

You might be interested in

What is the approximate area of a sector given O= 56 degrees with a diameter of 12m?

jeyben [28]

Area of sector is 17.584 meters

Solution:

Given that we have to find the approximate area of a sector given O= 56 degrees with a diameter of 12m

Diameter = 12 m

Radius = Diameter / 2 = 6 m

An angle of 56 degrees is the fraction $\frac{56}{360}$ of the whole rotation

A sector of a circle with a sector angle of 56 degrees is therefore also the fraction $\frac{56}{360}$ of the circle

The area of the sector will therefore also be $\frac{56}{360}$ of the area

$\text{ sector area } = \frac{56}{360} \times \pi r^2\\\\\text{ sector area } = \frac{56}{360} \times 3.14 \times 6^2\\\\\text{ sector area } = 17.584$

Thus area of sector is 17.584 meters

3 0

3 years ago

Answer pls fast FIND ALL ANSWERS

-BARSIC- [3]

Answer:

ebhats

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

I need the answer I will give brainliest answer !!!!!

Blizzard [7]

Answer:

Step-by-step explanation:

been struggling and crying I try my best and I still can’t I give up and please help me this is due today

6 0

3 years ago

Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot

Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

$\frac{dm}{dt} = -\frac{t}{\tau}$

Where:

$m$ - Current mass of the isotope, measured in kilograms.

$t$ - Time, measured in days.

$\tau$ - Time constant, measured in days.

The solution of the differential equation is:

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

Where $m_{o}$ is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

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

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

The half-life of a isotope ( $t_{1/2}$ ) as a function of time constant is:

$t_{1/2} = \tau \cdot \ln2$

$t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2$

The half-life difference between isotope B and isotope A is:

$\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|$

If $\frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9$ , $t_{A} = 33\,days$ and $t_{B} = 43\,days$ , the difference in the half-lives of the isotopes is:

$\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|$

$\Delta t_{1/2} \approx 65.788\,days$

The approximate difference in the half-lives of the isotopes is 66 days.

4 0

3 years ago

Read 2 more answers

I'm so confused!!!!! Can somebody plz help me!!!plz

ZanzabumX [31]

I would suggest looking up a converter chart, I don't know all of them but I can help with some.
For yards to feet you have to divide by 3 because there are 3 feet in a yard.
For lbs, it's 16 oz in a pound, so divide by 16.
For miles to yards you divide by 1760 because there are 1760 yards in a mile.
I hope this helps a little bit.

8 0

3 years ago