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3 years ago

Which of the following is equal to the product of 3 1/2 and 4/7 1/4 1/2 2 3.5

Mathematics

2 answers:

katovenus [111]3 years ago

7 0

I also got 2 hope this helps

zheka24 [161]3 years ago

4 0

Answer:

the answer is 2

Step-by-step explanation:

multiply 3 1/2 and 4/7 to hlget the product (2)

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Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

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3 years ago

A person invested $6,000, some at 7% and the rest at 5%. The income from the 7% investment is equal to the income from the 5% in

8090 [49]

Answer:

The person invested $3500 at 7% and $2500 at 5%

Step-by-step explanation:

We can say that $0,07x+0,05x=6000$ because the investment is equal both in one as in another, where x is the income from the investment.

Solving the equation we have:

$0,12x=6000\\x=50000$

Then to get the incomes:

At 7%: $I=0.07*50000\\I=$3500$

And at 5%: $I=0.05*50000\\I=$2500$

Finally we can demonstrate the answer because the income at 7% + income at 5% are 3500+2500=$6000

6 0

3 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago