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2 years ago

14. What is the product of -3x and 5x2 + x?

Mathematics

1 answer:

Volgvan2 years ago

5 0

Answer:

$-15x^{3} -3x^{2}$ OR $-x^{2} (15x+3)$

Step-by-step explanation:

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At the product store you can buy four bags of bananas for 2084 how much would it cost if you were to buy 3 bags

Darya [45]

Answer:

1563

Step-by-step explanation:

x= 1 bag of bananas

so, 4x=2084

Divide 2084 by 4 to find how much 1 bag of bananas would cost. 2084/4= 521.

To find how much 3 bags of bananas would cost, multiply 521 (the price of one bag) by 3. 521x3= 1563

3 bags of bananas are $1563

3 0

2 years ago

I don't under stand what is wrong.

MAXImum [283]

Answer:

In the first step, there is an issue when distributing the -3 to the terms inside of the parentheses. When multiplying the -3 to the -5, you would get an answer of +15, not -15.

The full correct process is:

(4m + 9) - 3(2m - 5) = 4m + 9 - 6m + 15

= 4m - 6m + 9 + 15

= -2m + 24

4 0

2 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

Which equation represents the polynomial function with zeros −1, 1, and 3 (multiplicity of 2), and a y-intercept of −18? (4 poin

Vlad [161]

Correct Answer:
3rd option is the correct answer

Solution:
The zeros of the polynomial are -1,1 and 3. The multiplicity of 3 is 2. So the polynomial can be expressed as:

$y=a (x-3)^{2}(x-1)(x+1)$

The y-intercept of the polynomial is -18. This means the polynomial passes through the point (0,-18). Therefore, y must be -18 when x = 0. Using these values of x and y in previous equation we get:

$-18=a (0-3)^{2}(0-1)(0+1) \\ \\ 
-18=-9a \\ \\ 
a=2$

The final equation of the polynomial becomes:

$y=2 (x-3)^{2}(x-1)(x+1)$

7 0

3 years ago

Solve (4x^4-7-6x^2) (2x^2 4x^4 8)

kotykmax [81]

Your answer can only be simplified to
256x^10 - 384^8 - 448^6

5 0

3 years ago