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liq [111]
3 years ago
9

307,407 rouned to the nereist thousandth

Mathematics
1 answer:
kobusy [5.1K]3 years ago
6 0
Rounded to the thousands would be 307,000 and thousandths is 307.407
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Gelneren [198K]
1 plus to the negative Peter plus the relative integers
3 0
3 years ago
What is proportional
AnnyKZ [126]

Answer: Proportional is two varying quantities are said to be in a relation of proportionality, if they are multiplicatively connected to a constant, that is, when either their ratio or their product yields a constant. The value of this constant is called the coefficient of proportionality or proportionality constant.

8 0
3 years ago
Multiply (x - 4)(x - 3)<br> PLZZZ get it RIGHT, SHOW YOUR WORK,AND NO LINKS PLZ!
Pepsi [2]

Answer:

b

Step-by-step explanation:

(x - 4) (x - 3)

x^2 - 3x - 4x + 12 (you distribute x in (x-4) to each of the terms x and -3 and multiply them. x*x is x^2 and x*(-3) is -3x. Then, you distribute -4 in (x-4) to  each of the terms x and -3 and multiply them. -4*x is -4x and -4*-3 is 12)

x^2 - 7x + 12 (B)

3 0
3 years ago
Read 2 more answers
Ryan has some nickels and dimes. altogether, he has 34 coins that total $2.60. write and solve a system of equations to determin
777dan777 [17]
N+d=34
.05n+.10d=2.60

Solve by substitution
n=34-d
.05(34-d)+.10d=2.60
1.7-.05d+.10d=2.60
.05d=.90
d=18 dimes
18+n=34
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4 0
3 years ago
Find an explicit solution of the given initial-value problem.dy/dx = ye^x^2, y(3) = 1.
tamaranim1 [39]

Answer:

y=exp(\int\limits^x_4 {e^{-t^{2} } } \, dt)

Step-by-step explanation:

This is a separable equation with an initial value i.e. y(3)=1.

Take y from right hand side and divide to left hand side ;Take dx from left hand side and multiply to right hand side:

\frac{dy}{y} =e^{-x^{2} }dx

Take t as a dummy variable, integrate both sides with respect to "t" and substituting x=t (e.g. dx=dt):

\int\limits^x_3 {\frac{1}{y} } \, \frac{dy}{dt} dt=\int\limits^x_3 {e^{-t^{2} } } dt

Integrate on both sides:

ln(y(t))\left \{ {{t=x} \atop {t=3}} \right. =\int\limits^x_3 {e^{-t^{2} } } \, dt

Use initial condition i.e. y(3) = 1:

ln(y(x))-(ln1)=\int\limits^x_3 {e^{-t^{2} } } \, dt\\ln(y(x))=\int\limits^x_3 {e^{-t^{2} } } \, dt\\

Taking exponents on both sides to remove "ln":

y=exp (\int\limits^x_3 {e^{-t^{2} } } \, dt)

7 0
3 years ago
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