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3 years ago

What is the solution to the system of equations?

Mathematics

1 answer:

Llana [10]3 years ago

3 0

Solution in attachment.

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Please help asap I suck at math :(

Lemur [1.5K]

Answer:

a. 68 mi/h

b. It will take Nolan 3 hours at this rate

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3 years ago

Read 2 more answers

Look at the image below

krok68 [10]

Answer:

SA = 153.9m^2

Step-by-step explanation:

SA = 4 $\pi$ $r^{2}$

r = 3.5

SA = 4 $\pi$ $(3.5)^{2}$

SA = 4 $\pi$ (12.25)

SA = 49 $\pi$

SA = 153.9m^2

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3 years ago

Bobby drives his truck at a rate of 50 to 60 miles per hour.

Yanka [14]

a. 2 hours = 100 to 120m

3 hrs = 150 to 180m

4 hrs = 200 to 240m

5 hrs = 250 to 300m

b. 400 to 480 miles

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3 years ago

Find the area of each circle to the nearest tenth. Use 3.14 for pi.

miv72 [106K]

Answer:

a=12.56 b=3.14 c=50.24 d=78.5 e=4.5216 f=63.585

Step-by-step explanation:

USE A CALCULATOR!

3 0

3 years ago

A) Find y' if x^3 + y^3 = 6xy.

olya-2409 [2.1K]

A)

$\bf x^3+y^3=6xy\\\\
-----------------------------\\\\
3x^2+3y^2\frac{dy}{dx}=6\left( 1\cdot y+x\frac{dy}{dx} \right)\\\\\\ 3\left( x^2+ y^2\frac{dy}{dx}\right)=6\left( 1\cdot y+x\frac{dy}{dx} \right)
\\\\\\
 x^2+ y^2\frac{dy}{dx}=2y+2x\frac{dy}{dx}\implies x^2+ y^2\frac{dy}{dx}-2y-2x\frac{dy}{dx}=0
\\\\\\
\cfrac{dy}{dx}(y^2-2x)=2y-x^2\implies \cfrac{dy}{dx}=\cfrac{2y-x^2}{y^2-2x}$

B)

$\bf \left. \cfrac{dy}{dx}=\cfrac{2y-x^2}{y^2-2x} \right|_{3,3}\implies 
\begin{cases}
x=3\\
y=3
\end{cases}\implies -1
\\\\
-----------------------------\\\\
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-3=-1(x-3)\\
\qquad \uparrow\\
\textit{point-slope form}$

solve for "y", and that'd be the equation of the tangent at 3,3

C) recall, the tangent is horizontal when the slope is 0, thus

$\bf \cfrac{dy}{dx}=\cfrac{2y-x^2}{y^2-2x}\implies 0=\cfrac{2y-x^2}{y^2-2x}\implies 0=2y-x^2\implies \cfrac{x^2}{2}=\boxed{y}
\\\\\\
now \qquad x^3+y^3=6xy\implies x^3+\left( \boxed{\cfrac{x^2}{2}} \right)^3=6x\left( \boxed{\cfrac{x^2}{2}} \right)
\\\\\\
x^3+\cfrac{x^6}{8}=3x^3\implies \cfrac{8x^3+x^6}{8}=3x^3
\\\\\\
8x^3+x^6-24x^3=0\implies x^6-16x^3=0
\\\\\\
x^3(x^2-16)=0\implies x=\{0,\pm 4\}$

so.. x =0 is not exactly in the first quadrant, x = -4 isn't either

so. the only choice is really x = 4... what's "y" when x = 4? well, just plug that into the $\bf x^3+y^3=6xy\qquad or \qquad \cfrac{x^2}{2}=y$

and solve for "y"

4 0

3 years ago