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3 years ago

9) work out 6 - 3x-2--10

Mathematics

1 answer:

Kaylis [27]3 years ago

3 0

3x+6 I believe because you can’t add a number with a variable to a number without one

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Find compound amount annually of p=4000,time=3/2 and rate =10% solve it step wise.

Natalka [10]

Step-by-step explanation:

Hey, there!!

Principal (p) = 4000

Time = 3/2= 1.5 yrs.

rate = 10%

Now, we have formula,

$c.a = p \times {(1 + \frac{r}{100}) }^{t}$

Putting their values,

$ca = 4000\times {(1 + \frac{10}{100} )}^{1.5}$

$ca =4000 \times {(1 + 0.1)}^{1.5}$

$ca = 4000 \times 1.153689$

Simplifying them we get,

C.A = 4614.75

Hope it helps...

4 0

3 years ago

What is equivalent to -7/-9

docker41 [41]

-7/-9 is equivalent to -14/-18. Multiply the denominator and numerator by the same number(I used 2)

3 0

3 years ago

A car manufacturer sold about 332,000 cars last year. The number of cars

Viefleur [7K]

Answer:

415,000

Step-by-step explanation:

First, find the decimal equivalent of 125%. To do this, divide 125 by 100 or move two decimal places to the right: 125/100 = 1.25

Now multiply the amount of cars sold last year, 332,000, by the percentage decimal, 1.25: 332,000 x 1.25 = 415,000

The car manufacturer sold 415,000 cars this year

4 0

4 years ago

Please help I don't have enough brain cells for this ;-;

bonufazy [111]

Answer:

$\frac{27}{22}$

Step-by-step explanation:

4 0

3 years ago

X ^ (2) y '' - 7xy '+ 16y = 0, y1 = x ^ 4

nignag [31]

Standard reduction of order procedure: suppose there is a second solution of the form $y_2(x)=v(x)y_1(x)$ , which has derivatives

$y_2=vx^4$
${y_2}'=v'x^4+4vx^3$
${y_2}''=v''x^4+8v'x^3+12vx^2$

Substitute these terms into the ODE:

$x^2(v''x^4+8v'x^3+12vx^2)-7x(v'x^4+4vx^3)+16vx^4=0$
$v''x^6+8v'x^5+12vx^4-7v'x^5-28vx^4+16vx^4=0$
$v''x^6+v'x^5=0$

and replacing $v'=w$ , we have an ODE linear in $w$ :

$w'x^6+wx^5=0$

Divide both sides by $x^5$ , giving

$w'x+w=0$

and noting that the left hand side is a derivative of a product, namely

$\dfrac{\mathrm d}{\mathrm dx}[wx]=0$

we can then integrate both sides to obtain

$wx=C_1$
$w=\dfrac{C_1}x$

Solve for $v$ :

$v'=\dfrac{C_1}x$
$v=C_1\ln|x|+C_2$

Now

$y=C_1x^4\ln|x|+C_2x^4$

where the second term is already accounted for by $y_1$ , which means $y_2=x^4\ln x$ , and the above is the general solution for the ODE.

4 0

3 years ago

9) work out 6 - 3x-2--10​

9) work out 6 - 3x-2--10