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IrinaVladis [17]
3 years ago
6

The population of a city is 2,500. If the number of males is 240 more than the

Mathematics
1 answer:
klio [65]3 years ago
7 0
That means there are:
1,130 females
&
1,370 males.
You might be interested in
Your bank account has a balance of $45.24.this is $51.35 more than yesterday. What was your account balance yesterday?
Nezavi [6.7K]

Answer: -6.11

Step-by-step explanation:

Subtract.

 45.24

- 51.35

----------------

Invert the operation because since the result is a negative number, we cannot operate 51 of 45. Just keep in mind that the result is going to be negative because 51 is bigger than 45.

  51.35

- 45.24

-------------

   -6.11

5 0
3 years ago
Read 2 more answers
What is the answer?
Dmitrij [34]

Answer:

read explanation

Step-by-step explanation:

rearrange equations to form y=mx+b

-2x+4=2y

-x+2=y

-y=-2x-5

y=2x+5

use these equations to find slope and y intercepts, and match those with the right graph

slope of first: -1

slope of second: 2

y-int of first: 2

y-int of second:5

3 0
3 years ago
5 p increased by 25% is equal to q decreased by 25%. Work out p as a percentage of q.​
oksano4ka [1.4K]

The answer is no it will be less here's why let say the original amount is A then a p% increase will be A(1+p/100))

Now from this amount we decrease it by p% this gives A(1+p/100)(1-p/100). We can simplify this expression some by letting x=p/100 then we have A(1+x)(1-x) but this is equal to A(1-x2). You can see that the value of what's inside the () is always <1 as long as x>0. If we calculate how much the amount changes we have A(1-x2)-A=-Ax2 is always a loss.

         This is one of the reasons some people lose money in the stock market. If a stock rises 10% and then drops 10% you have actually lost money but it sounds, on the face of it, like you should be even. +10 and -10 percent.

          If we now do the calculation for the 25% up and down we have the final amount is is .9375 which is 1.25*.75. and the change is .9375-1=-.0625. So the net result is down 6.25%

6 0
2 years ago
Read 2 more answers
A mass weighing 16 pounds stretches a spring (8/3) feet. The mass is initially released from rest from a point 2 feet below the
mezya [45]

Answer with Step-by-step explanation:

Let a mass weighing 16 pounds stretches a spring \frac{8}{3} feet.

Mass=m=\frac{W}{g}

Mass=m=\frac{16}{32}

g=32 ft/s^2

Mass,m=\frac{1}{2} Slug

By hook's law

w=kx

16=\frac{8}{3} k

k=\frac{16\times 3}{8}=6 lb/ft

f(t)=10cos(3t)

A damping force is numerically equal to 1/2 the instantaneous velocity

\beta=\frac{1}{2}

Equation of motion :

m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}+f(t)

Using this equation

\frac{1}{2}\frac{d^2x}{dt^2}=-6x-\frac{1}{2}\frac{dx}{dt}+10cos(3t)

\frac{1}{2}\frac{d^2x}{dt^2}+\frac{1}{2}\frac{dx}{dt}+6x=10cos(3t)

\frac{d^2x}{dt^2}+\frac{dx}{dt}+12x=20cos(3t)

Auxillary equation

m^2+m+12=0

m=\frac{-1\pm\sqrt{1-4(1)(12)}}{2}

m=\frac{-1\pmi\sqrt{47}}{2}

m_1=\frac{-1+i\sqrt{47}}{2}

m_2=\frac{-1-i\sqrt{47}}{2}

Complementary function

e^{\frac{-t}{2}}(c_1cos\frac{\sqrt{47}}{2}+c_2sin\frac{\sqrt{47}}{2})

To find the particular solution using undetermined coefficient method

x_p(t)=Acos(3t)+Bsin(3t)

x'_p(t)=-3Asin(3t)+3Bcos(3t)

x''_p(t)=-9Acos(3t)-9sin(3t)

This solution satisfied the equation therefore, substitute the values in the differential equation

-9Acos(3t)-9Bsin(3t)-3Asin(3t)+3Bcos(3t)+12(Acos(3t)+Bsin(3t))=20cos(3t)

(3B+3A)cos(3t)+(3B-3A)sin(3t)=20cso(3t)

Comparing on both sides

3B+3A=20

3B-3A=0

Adding both equation then, we get

6B=20

B=\frac{20}{6}=\frac{10}{3}

Substitute the value of B in any equation

3A+10=20

3A=20-10=10

A=\frac{10}{3}

Particular solution, x_p(t)=\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)

Now, the general solution

x(t)=e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)

From initial condition

x(0)=2 ft

x'(0)=0

Substitute the values t=0 and x(0)=2

2=c_1+\frac{10}{3}

2-\frac{10}{3}=c_1

c_1=\frac{-4}{3}

x'(t)=-\frac{1}{2}e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+e^{-\frac{t}{2}}(-c_1\frac{\sqrt{47}}{2}sin(\frac{\sqrt{47}t}{2})+\frac{\sqrt{47}}{2}c_2cos(\frac{\sqrt{47}t}{2})-10sin(3t)+10cos(3t)

Substitute x'(0)=0

0=-\frac{1}{2}\times c_1+10+\frac{\sqrt{47}}{2}c_2

\frac{\sqrt{47}}{2}c_2-\frac{1}{2}\times \frac{-4}{3}+10=0

\frac{\sqrt{47}}{2}c_2=-\frac{2}{3}-10=-\frac{32}{3}

c_2==-\frac{64}{3\sqrt{47}}

Substitute the values then we get

x(t)=e^{-\frac{t}{2}}(-\frac{4}{3}cos(\frac{\sqrt{47}t}{2})-\frac{64}{3\sqrt{47}}sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)

8 0
4 years ago
Equivalent expression of 7 square root 5
Vika [28.1K]

Answer:19

Step-by-step explanation:

Subtract then add

4 0
3 years ago
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