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serg [7]
3 years ago
7

Question 5

Mathematics
1 answer:
kumpel [21]3 years ago
8 0

Answer:

Congruent

Explanation:

The reflection never changes the angle measures or side lengths of the figure.

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Flying to Kimpala with a tail wind a plane averaged 158km/hr. On the return trip the plane only average 112km/hr. While flying b
lora16 [44]

The speed of wind and speed of plane in still air  are  23  and  135 km/h respectively.

<u>Step-by-step explanation:</u>

Let the speed of wind and speed of plane in still air  are  w  and  p km/h respectively.  

The effective speed on onward journey was  

p+w=158  ................(1)

The effective speed on return journey was  

p-w=112 ..............(2)

Adding equation (1) and equation (2) we get,  

⇒ (p+w)+(p-w) = 158+112

⇒ 2p = 270

⇒ p = 135

Putting value of p = 135 in p+w=158 we get:

⇒ p+w=158

⇒ 135+w=158

⇒ w=23

Therefore ,The speed of wind and speed of plane in still air  are  23  and  135 km/h respectively.

3 0
3 years ago
A fishing tackle box is 13 inches long, 6 inches wide, and two and one half inches high. What is the volume of the tackle box
weqwewe [10]

Answer:

195^{3} inches

Step-by-step explanation:

13*6*2.5

7 0
3 years ago
Which of the following situations could be modeled by an exponential function?
Sauron [17]

Answer: Option D.

Step-by-step explanation:

An exponential function is something like:

f(x) = A*(r)^x

Where:

A = initial amount.

r = rate of growth

x = variable.

This type of function is used to describe situations like "you have A dollars, and it increases by 30% each week"

or:

"There is a population of A animals, and it doubles each year"

Let's analyze briefly all the given options and see which one is an exponential function.

A) The water level of a tank decreases by 5 gallons every day.

This is a linear relation, if W is the initial amount of water, and x is the number of days, then the water level of the tank after x days is:

y = W - 5 gal*x

B) The amount of money collected by a charity increases by $5,000 each year.

The amount collected increases $5000 each year.

Then if in year 1 they collected A.

in year 2 they collected A + $5000

in year 3 they collected A + 2*$5000

We already can see the pattern, this is a linear relation:

in year X they colected A + (X - 1)*$5000

C) A tree was 10 feet tall when purchased and grows at a rate of 2 feet per year.

if y is the height of the tree, and x is the number of years, this can be written as:

y = 10ft + x*2ft

Again, a linear relation.

D) The number of people in a stadium is decreasing at a rate of 45% every 10 minutes.

Let's define x as each "lapse" of 10 minutes that passes.

If N is the initial number of people in te stadium, then after 10 minutes the number of people is = N*(1 - 45%/100%) = N*(1 - 0.45) = N*0.55

After another 10 minutes, the population decreases again.

N*0.55*(1 - 45%/100%) = N*0.55*0.55 = N*0.55^2

We can see the pattern here, this can be written as

People(x) = N*(0.55)^x

This is an exponential function.

3 0
2 years ago
Suppose that the bacteria in a colony grow unchecked according to the Law of Exponential Change. The colony starts with 1 bacter
inn [45]

Answer: There are 2.25×10³⁴ bacteria at the end of 24 hours.

Step-by-step explanation:

Since we have given that

Number of bacteria initially = 1

It triples in number every 20 minutes.

So, \dfrac{20}{60}=\dfrac{1}{3}

So, our equation becomes

y=y_0e^{\frac{1}{3}k}\\\\3=1e^{\frac{1}{3}k}\\\\\ln 3=\dfrac{1}{3}k\\\\k=\dfrac{1.099}{0.333}=3.3

We need to find the number of bacteria that it will contain at the end of 24 hours.

So, it becomes,

y=1e^{24\times 3.3}\\\\y=e^{79.1}\\\\y=2.25\times 10^{34}

Hence, there are 2.25×10³⁴ bacteria at the end of 24 hours.

3 0
3 years ago
Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2
igor_vitrenko [27]

Answer:  The required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Step-by-step explanation:  We are given to find the solution of the following initial value problem :

y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.

Let y=e^{mx} be an auxiliary solution of the given differential equation.

Then, we have

y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.

Substituting these values in the given differential equation, we have

m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.

So, the general solution of the given equation is

y(x)=Ae^x+Be^{-x}, where A and B are constants.

This gives, after differentiating with respect to x that

y^\prime(x)=Ae^x-Be^{-x}.

The given conditions implies that

y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

and

y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)

Adding equations (i) and (ii), we get

2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.

From equation (i), we get

\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.

Substituting the values of A and B in the general solution, we get

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Thus, the required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

4 0
3 years ago
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