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3 years ago

Solve for x, given the equation Square root of x-5+7=11

Mathematics

1 answer:

denis-greek [22]3 years ago

6 0

$\sqrt{x-5+7} = 11\\x-5+7 = 11^2\\x-5+7 = 121\\x + 2 = 121\\x = 119$

Check the answer:

$\sqrt{119-5+7} \\\sqrt{114+7} \\\sqrt{121} \\11$

This answer is correct,

x = 119

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A savings account accrues interest at a rate of 3.0% yearly. If someone opens an account with $2,500, how much money would the a

stira [4]

<span>2500(1.03)^5 that's the equation .
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6 0

3 years ago

The bike store marks up the wholesale cost of all of the bikes they sell by 30%. Andre wants to buy a bike that has a price tag

Step2247 [10]

Answer:

$96.15

Step-by-step explanation:

Given data

markup = 30%

cost price= $125

let the cost price before markup be x

125-30/100*x= x

125-0.3x= x

125= x+0.3x

125= 1.3x

divide both sides by 1.3

x= 125/1.3

x= $96.15

Hence the price before markup is $96.15

6 0

3 years ago

Help please <br> I need to pass my class

valina [46]

Answer:

look below

Step-by-step explanation:

thing to note:

sin = opposite/hypotenuse

cos = adjacent/hypotenuse

tan = opposite/adjacent

opposite= side across angle you are using

hypotenuse= side across 90 degree angle

adjacent= remaining side

angle B

sin = 6/10

cos = 8/10

tan = 6/8

angle C

sin = 8/10

cos = 6/10

tan = 8/6

hope that helped. let me know if you have any questions:)

3 0

3 years ago

Find the output of the function y = x - 16 if the input is -28.<br> Help please

Serhud [2]

Answer:

-44

Step-by-step explanation:

If the input is -28, that means x = -28, so we plug -28 in for x and solve for y, which is the output:

y = x - 16

y = -28 - 16 = -44

The answer is thus -44.

5 0

3 years ago

Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)

babunello [35]

$\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k$

We want to find $f(x,y,z)$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=x^2+y$

$\dfrac{\partial f}{\partial y}=y^2+x$

$\dfrac{\partial f}{\partial z}=ze^z$

Integrating both sides of the latter equation with respect to $z$ tells us

$f(x,y,z)=e^z(z-1)+g(x,y)$

and differentiating with respect to $x$ gives

$x^2+y=\dfrac{\partial g}{\partial x}$

Integrating both sides with respect to $x$ gives

$g(x,y)=\dfrac{x^3}3+xy+h(y)$

Then

$f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)$

and differentiating both sides with respect to $y$ gives

$y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C$

So the scalar potential function is

$\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}$

By the fundamental theorem of calculus, the work done by $\vec F$ along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it $L$ ) in part (a) is

$\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}$

and $\vec F$ does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them $L_1$ and $L_2$ ) of the given path. Using the fundamental theorem makes this trivial:

$\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3$

$\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4$

8 0

3 years ago