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3 years ago

I am a mutiple of 60. I am between 400 and 600 . I am 40 away from a multiple of 100. who am i

1 answer:

3 0

There is no possible answer.
The #s are 420, 480, and 540.
None of them are 40 away from a multiple of 100.
However, it is was between 300 and 600, the answer would be 360.

You might be interested in

Is 23 a prime number

Gekata [30.6K]

Answer:

Step-by-step explanation:

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. A natural number greater than 1 that is not prime is called a composite number.

3 0

3 years ago

What is the x-intercept and the y-intercept for x+2y=8

Ierofanga [76]

To find the x-intercept, you need to set y equal to zero(think about this on a graph!)
This will become:
x + 2(0) = 8
If we remove the unnecessary zero:
x = 8
That's the x intercept, which can be expressed as the point (8,0).

To find the y-intercept, you need to set x equal to zero(again, think about that on a graph!)
This becomes:
0 + 2y = 8
Remove the unnecessary 0:
2y = 8
Divide both sides by 2:
y = 4
There ya go! Or, in point form: (0, 4)

Hope this helped! :)
~Chrys

8 0

3 years ago

Which ordered pairs are solutions to the inequality −2x+y≥−4 ?

GalinKa [24]

The answer would be A

3 0

3 years ago

david quiere saber cuanto requiere invertir mensualmente para obtener $30,760.08 durante 6 meses si los invierte con el 12% capi

Akimi4 [234]

Answer:

A = $32,652.44

Step-by-step explanation:

Given: Principal (P) = 30,760.08, Annual Rate (R) = 12%, Time (t in years) = 0.5

To find: How much David needs to invest monthly

Formula: $A = P(1 + r/n)^n^t$

Solution: To find, simply add principal + interest

First, convert R as a percent to r as a decimal

r = R/100

r = 12/100

r = 0.12 rate per year,

Then solve the equation for A

A = P(1 + r/n)nt

A = 30,760.08(1 + 0.12/12)(12)(0.5)

A = 30,760.08(1 + 0.01)(6)

A = $32,652.44

Therefore;

The total amount David will obtain with 30,760.08 for 6 months, 12%is $32,652.44.

4 0

3 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

3 years ago