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djverab [1.8K]
3 years ago
13

What is the area, in square meters, of the shape below? Express your answer as a fraction in simplest form.

Mathematics
1 answer:
kramer3 years ago
5 0

Answer:

20/3

Step-by-step explanation:

5/3 x 4/3

5x4=20

keep the denominator

20/3

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Need help with this.
skelet666 [1.2K]

Answer:

-5/2(3x+4)<6-3x (multiply with -5/2)

-15/2x-10<6-3x (multiply with 2)

-15x-20<12-6x (change sides)

-15x+6x<12+20

-9x<12+20

-9x<32

x>-32/9

Hope this will help u :)

3 0
4 years ago
What is the gradient of the line<br> y = 3 - 5x?
arsen [322]
-5

u can rearrange it to be y = -5x + 3
4 0
3 years ago
Help with trigonometry hw
DochEvi [55]

Step-by-step explanation:

Remember:

SOH - Sin(angle) = Opposite/Hypotenuse

CAH - Cos(angle) = Adjacent/Hypotenuse

TOA - Tan(angle) = Opposite/Adjacent

3 0
3 years ago
Suppose an unknown radioactive substance produces 8000 counts per minute on a Geiger counter at a certain time, and only 500 cou
mariarad [96]

Answer:

The half-life of the radioactive substance is of 3.25 days.

Step-by-step explanation:

The amount of radioactive substance is proportional to the number of counts per minute:

This means that the amount is given by the following differential equation:

\frac{dQ}{dt} = -kQ

In which k is the decay rate.

The solution is:

Q(t) = Q(0)e^{-kt}

In which Q(0) is the initial amount:

8000 counts per minute on a Geiger counter at a certain time

This means that Q(0) = 8000

500 counts per minute 13 days later.

This means that Q(13) = 500. We use this to find k.

Q(t) = Q(0)e^{-kt}

500 = 8000e^{-13k}

e^{-13k} = \frac{500}{8000}

\ln{e^{-13k}} = \ln{\frac{500}{8000}}

-13k = \ln{\frac{500}{8000}}

k = -\frac{\ln{\frac{500}{8000}}}{13}

k = 0.2133

So

Q(t) = Q(0)e^{-0.2133t}

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

Q(t) = Q(0)e^{-0.2133t}

0.5Q(0) = Q(0)e^{-0.2133t}

e^{-0.2133t} = 0.5

\ln{e^{-0.2133t}} = \ln{0.5}

-0.2133t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.2133}

t = 3.25

The half-life of the radioactive substance is of 3.25 days.

7 0
3 years ago
Help with problem pls
Crazy boy [7]
The original number is .9
5 0
4 years ago
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