Question 4 of 5

Mathematics

2 answers:

sergejj [24]4 years ago

6 0

Answer:

C. A'(-5-5), B'(-1,-5), C (-1.-2) D'(-5-1)

Step-by-step explanation:

Your welcome

grigory [225]4 years ago

3 0

It will definitely be C

any time there it a reflection across the x-axis the coordinates then become (x,-y)

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Y=6x4 + 7x2 and x = vw +1

belka [17]

Answer:

w=-1/6 or w=-3

Step-by-step explanation:

Substitute the value of x and y into the first equation and solve for w:

$y=6x^4+7x^2\\10=6(\sqrt{w+1})^4+7(\sqrt{w+1})^2\\10=6(w+1)^2+7(w+1)\\10=6(w^2+2w+1)+7w+7\\10=6w^2+12w+6+7w+7\\0=6w^2+19w+3$

Solve for w with the quadratic formula:

$w=\frac{-b\±\sqrt{b^2-4ac}}{2a} \\w=\frac{-19\±\sqrt{19^2-4(6)(3)} }{2(6)}\\w=\frac{-19\±\sqrt{361-72}}{12} \\w=\frac{-19\±\sqrt{289} }{12} \\w=\frac{-19\±17}{12}$

After simplifying, we get either w=-1/6 or w=-3.

Not sure if w needs to satisfy all conditions while maintaining everything as real numbers, because if w=-3, then x is an imaginary number. But both values work just as fine.

4 0

2 years ago

Find the volume of revolution bounded by the curves y = 4 – x2 , y = x, and x = 0, and is revolved about the vertical axis.

aleksley [76]

$4-x^2=x\\
x^2+x-4=0\\
\Delta=1^2-4\cdot1\cdot(-4)=1+16=17\\
x_1=\dfrac{-1-\sqrt{17}}{2}\\
x_2=\dfrac{-1+\sqrt{17}}{2}\\\\
\displaystyle
V=\pi\int\limits_0^{\dfrac{-1+\sqrt{17}}{2}}(4-x^2-x)^2\,dx\\
V=\pi\int\limits_0^{\dfrac{-1+\sqrt{17}}{2}}(16-4x^2-4x-4x^2+x^4+x^3-4x+x^3+x^2)\,dx\\
V=\pi\int\limits_0^{\dfrac{-1+\sqrt{17}}{2}}(x^4+2x^3-7x^2-8x+16)\,dx\\
V=\pi \left[\dfrac{x^5}{5}+\dfrac{x^4}{2}-\dfrac{7x^3}{3}-4x^2+16x\right]_0^{\dfrac{-1+\sqrt{17}}{2}}\\
$

The rest of solution in the attachment.

There's a mistake in the picture
It shoud be
$V=\pi\left(\dfrac{289\sqrt{17}-521}{60}\right)\approx35$