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yanalaym [24]
3 years ago
10

Can someone please help me!

Mathematics
1 answer:
Tresset [83]3 years ago
6 0

1, 2, 3, 4, 5, 6, 7, 8, i don't know, and i haven't ate, 9, 10, 11, 12, this really sucks cuz ur on ur own :(

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Interest = $45.58, time = 1 year, rate = 2.25%. Find the principal. Round to the nearest hundredth place
Murrr4er [49]

Answer:

<em>$2,025.78 </em>

Step-by-step explanation:

I = Prt ⇒ P = \frac{I}{rt}

2.25% = 0.0225

P = 45.58 ÷ ( 0.0225 × 1 ) ≈ <em>$2,025.78</em>

8 0
2 years ago
What is the y-intercept of the function f(x) = –2/9x + 1/3?
marishachu [46]
<span>f(x) = –2/9x + 1/3
</span>y -intercept when x = 0
so
f(x) = –2/9(0) + 1/3 then f(x) = 1/3

answer
<span>y-intercept </span>(0, 1/3)
4 0
3 years ago
Read 2 more answers
What is the value of x? 6(-3-x)-2x=14
Lera25 [3.4K]
The equation given in the question has one unknown variable in the form of "x" and only one equation is required to find the value of the unknown variable. So it is absolutely possible to find the exact value of "x".
6(-3 - x) - 2x = 14
-18 - 6x - 2x = 14
-18 - 8x = 14
Multiplying both sides of the equation by -1 we get
8x + 18 = - 14
Dividing both sides by 2 we get
4x + 9 = - 7
4x = - 7 - 9
4x = - 16
x = - 16/4
  = - 4
So the value of "x" comes out to be equal to - 4. I hope the procedure is clear enough for you to understand.
7 0
3 years ago
Read 2 more answers
What is the value of log^381?A.2B.3C.4D.5
Veseljchak [2.6K]

Answer:

4

Step-by-step explanation:

7 0
3 years ago
Find a solution to the following initial-value problem: dy dx = y(y − 2)e x , y (0) = 1.
Veseljchak [2.6K]

This equation is separable, as

\dfrac{\mathrm dy}{\mathrm dx}=y(y-2)e^x\implies\dfrac{\mathrm dy}{y(y-2)}=e^x\,\mathrm dx

Integrate both sides; on the left, expand the fraction as

\dfrac1{y(y-2)}=\dfrac12\left(\dfrac1{y-2}-\dfrac1y\right)

Then

\displaystyle\int\frac{\mathrm dy}{y(y-2)}=\int e^x\,\mathrm dx\implies\frac12(\ln|y-2|-\ln|y|)=e^x+C

\implies\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x+C

Since y(0)=1, we get

\dfrac12\ln\left|\dfrac{1-2}1\right|=e^0+C\implies C=-1

so that the particular solution is

\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x-1\implies\boxed{y=\dfrac2{1-e^{2e^x-2}}}

4 0
3 years ago
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