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3 years ago

2x+1=5x-20what is the value of x?

Mathematics

2 answers:

goldenfox [79]3 years ago

8 0

$2x+1=5x-20\\
3x=21\\
x=7$

SIZIF [17.4K]3 years ago

7 0

2x+1=5x-20
you want to find the x value so you want to isolate x to do this you have to subtract 2x on both sides. After you have accomplished this you add 20 on both sides once again and then you divide what your answer from what you added by three.

2x+1=5x-20
-2x -2x
________
1=3x-20
+20 +20
_______
21/3=3x/3
7=x

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Step-by-step explanation:

P(t) = 800t - t²

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3 years ago

Jerry says I've got my money in a great account that compounds interest monthly. The equation y=388 (1.008) represents how much

zheka24 [161]

Answer:

(a)His monthly Interest Rate=0.8%

(b)Annual Interest Rate = 9.6%

Step-by-step explanation:

For a Principal P invested at a yearly rate r, compounded m times in t years

Amount at Compound Interest= $P(1+\frac{r}{m})^{mt}$

Comparing with Jerry's equation y=388 (1.008)

(a)His monthly Interest Rate= 0.008=0.8%

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(c)If I invest $500 at the same rate of return,

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= $P(1+\frac{r}{m})^{mt}$ $=500(1+0.008)^{m}$ $=500(1.008)^m$

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4 years ago

Evaluate the function for x=-5 F(x)=-6+7

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3 years ago

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Find the value ofStartFraction partial derivative z Over partial derivative x EndFraction∂z∂xFind the value of at the point (1,1

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Answer:

The value of $\frac{\partial z}{\partial x}$ at (1,1,1) is 1.

Step-by-step explanation:

Given,

$z''x+8xy-2yz-7=0\hfill (1)$

wnere, z=f(x,y).

We have to find,

$\frac{\partial z}{\partial x}$

From (1) letting z'=m we get,

$m^2-2ym+(8xy-7)=0$

$\implies m=\frac{-(-2y)\pm \sqrt{(-2y)^2-4\times(8xy-7)}}{2}$

$=\frac{2y\pm \sqrt{4y^2-8xy+7}}{2}$

$\therefore m=z'=\frac{dz}{dx}=y\pm \sqrt{y^2=8xy+7}$

$dz=y dx\pm \sqrt{y^2-8xy+7}dx$

Integrate by letting $u=y^2-8xy+7$ we get,

$\int dz=xy\pm \int \frac{\sqrt{u}}{8y}$

$z(x,y)=xy\pm \frac{1}{8y}\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}$

$=xy\pm\frac{(y^2-8xy+7)^{\frac{5}{2}}}{12y}$

Now differentiate partially with respect to x we get,

$\frac{\partial z}{\partial x}=y\pm\frac{\partial}{\partial x}\frac{(y^2-8xy+7)^{\frac{5}{2}}}{12y}$

$=y\pm\frac{1}{12y}\frac{\partial}{\partial x}(y^2-8xy+7)^{\frac{5}{2}$

Letting $u=y^2-8xy+7, f=u^\frac{5}{2}$ and applying chain rule,

$\frac{df(u)}{dx}=\frac{df}{du}\frac{du}{dx}$ we get,

$\frac{\partial z}{\partial x}=y\pm\frac{1}{12y}\frac{\partial}{\partial u}u^{\frac{5}{2}}\frac{\partial}{\partial x}(y^2-8xy+7)$

$=y\pm\frac{5}{3}(y^2-8xy+7)^{\frac{3}{2}$

At (1,1,1),

$\Big[\frac{\partial z}{\partial x}\Big]_{(1,1,1)}=1$

Hence the result.

7 0

4 years ago