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Blizzard [7]
3 years ago
15

Triangle OPQ is similar to triangle PST. Need this asap

Mathematics
1 answer:
mars1129 [50]3 years ago
3 0

Answer:

58.2

Step-by-step explanation:

We can find the length of RS using similarity ratio

7/34 = 12/RS cross multiply expressions

7RS = 408 divide both sides by 7

RS = 58.2 approximately.

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What is the domain and range for the following function and its inverse?
Basile [38]

Answer:

"f(x)  

domain: all real numbers, range: all real numbers  

f–1(x)  

domain: all real numbers, range: all real numbers"

Step-by-step explanation:

We can use the fact that the domain of a function and the range of its inverse are equal.

Also, the range of the function and the domain of its inverse are equal as well.

<em>Looking at the function f(x/ = -x + 5, we see that this is a line with a negative slope of 1 and a y-intercept of +5. </em>

As we know from the graph of lines, there is no restricting values in x and y. So for the original function,  domain is the set of all real numbers and the range is the set of all real numbers.

For the inverse, the range is set of all real numbers and domain is also the set of all real numbers.

First answer choice is right.

4 0
3 years ago
Four circles, each with a radius of 2 inches, are removed from a square. What is the remaining area of the square? (16 – 4π) in.
anyanavicka [17]
Since 4 circles are circumscribed by a square, then the side length of the square is 8

So the area of the square is 8^2 = 64

The area of 4 squares is 4 * pi*2^2 = 16pi

So the answer will be 64 - 16pi
6 0
3 years ago
Read 2 more answers
Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca
Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

\frac{dQ}{dt} = -rt

Which has the following solution:

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

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that Q(6) = (1-0.03)Q(0) = 0.97Q(0). We use this to find r.

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

0.97Q(0) = Q(0)e^{-6r}

e^{-6r} = 0.97

\ln{e^{-6r}} = \ln{0.97}

-6r = \ln{0.97}

r = -\frac{\ln{0.97}}{6}

r = 0.0051

So

Q(t) = Q(0)e^{-0.0051t}

Determine the half-life of the radioactive substance.

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

Q(t) = Q(0)e^{-0.0051t}

0.5Q(0) = Q(0)e^{-0.0051t}

e^{-0.0051t} = 0.5

\ln{e^{-0.0051t}} = \ln{0.5}

-0.0051t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.0051}

t = 135.9

The half-life of the radioactive substance is 135.9 hours.

6 0
3 years ago
A = 40, c = 41, b =<br><br> help please
nikitadnepr [17]

Answer:

I'm thinking it's 42. Did you notice that there is a pattern?

4 0
3 years ago
Part B: Solve 2a − 5d = 30 for d. Show your work. (6 points)
stich3 [128]
<span>Given Equation: 2a − 5d = 30 

Solution:

2a - 5d = 30

-5d = 30 -2a

d = ( 30 - 2a )/ (-5)

  =  -6 + (2a/5) 

  = (2a/5) - 6

I hope this helps</span>
3 0
4 years ago
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