What os the correct question and why

) Enter a negative number that has an absolute value greater than 10.

Mkey [24]

Answer:

-11

Step-by-step explanation:

5 0

3 years ago

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PLEASE HELP

laila [671]

<h3><u>Option A</u></h3>

$4 x^{2}+32 x-80=0$ is the required equation to calculate width of rectangular frame that has a total area of 140 square inches.

<h3><u>Solution:</u></h3>

Given that,

Length of a rectangular frame is given as 2x + 10

Width of the rectangular frame is given as 2x + 6

Total area = 140 square inches

<em><u>The area of rectangular frame is given as:</u></em>

$\text {area of rectangle}=\text {length } \times \text {width}$

Plugging in values, we get

$\begin{array}{l}{(2 x+10)(2 x+6)=140} \\\\ {4 x^{2}+12 x+20 x=80} \\\\ {4 x^{2}+32 x-80=0}\end{array}$

This is the required equation to calculate width of rectangular frame

Solve the above quadratic equation to get the value of "x"

$4 x^{2}+32 x-80=0$

<em><u>Use the quadratic equation formula:</u></em>

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here a = 4 ; b = 32 ; c = -80

$x=\frac{-32 \pm \sqrt{32^{2}-4(4)(-80)}}{2(4)}$

$x=\frac{-32 \pm \sqrt{2304}}{8}$

x = 2 or x = -10

Now measurement cannot be negative, so taking the positve value of "x", we can calculate the width

So put "x" = 2

Width of the rectangular frame = 2x + 6 = 2(2) + 6 = 10

Thus the width of frame is 10 inches

4 0

3 years ago

Find the missing lengths of the sides.

ELEN [110]

Answer:

C:a = 9 in., b = in.

Step-by-step explanation:

The square root of 2 is added on the hypotenuse of a triangle with the angles 45-45-90 while(in this case) a and b are equal

3 0

2 years ago

Find the general solution of the equation: (y^2-2xy)dx + (2xy-x^2)dy = 0

lys-0071 [83]

The ODE is exact, since

$\dfrac{\partial(y^2-2xy)}{\partial y}=2y-2x$

$\dfrac{\partial(2xy-x^2)}{\partial x}=2y-2x$

so there is solution $f(x,y)=C$ such that

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

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

Integrating both sides of the first PDE wrt $x$ gives

$f(x,y)=xy^2-x^2y+g(y)$

Differentiating both sides wrt $y$ gives

$\dfrac{\partial f}{\partial y}=2xy-x^2+\dfrac{\mathrm dg}{\mathrm dy}=2xy-x^2$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=0\implies g(y)=C$

Then the solution to the ODE is

$f(x,y)=\boxed{xy^2-x^2y=C}$

# # #

Alternatively, we can see that the ODE is homogeneous, since replacing $x\to tx$ and $y\to ty$ reduces to the same ODE:

$((ty)^2-2(tx)(ty))\,\mathrm d(tx)+(2(tx)(ty)-(tx)^2)\,\mathrm d(ty)=0$

$t^3(y^2-2xy)\,\mathrm dx+t^3(2xy-x^2)\,\mathrm dy=0$

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

This tells us we can solve by substituting $y(x)=xv(x)$ , so that $\mathrm dy(x)=x\,\mathrm dv(x)+v(x)\,\mathrm dx$ , and the ODE becomes

$(x^2v^2-2x^2v)\,\mathrm dx+(2x^2v-x^2)(x\,\mathrm dv+v\,\mathrm dx)=0$

$v(v-2)\,\mathrm dx+(2v-1)(x\,\mathrm dv+v\,\mathrm dx)=0$

$3v(v-1)\,\mathrm dx+x(2v-1)\,\mathrm dv=0$

which is separable as

$\dfrac{1-2v}{3v(v-1)}\,\mathrm dv=\dfrac{\mathrm dx}x$

Integrating both sides gives

$-\dfrac13\ln|v(1-v)|=\ln|x|+C$

$v(1-v)=\dfrac C{x^3}$

and solving in terms of $y(x)$ ,

$\dfrac yx\left(1-\dfrac yx\right)=\dfrac C{x^3}$

$xy(x-y)=C$

$\boxed{x^2y-xy^2=C}$

6 0

2 years ago

When triangle ABC is rotated about side AB, what figure is formed?

Pani-rosa [81]

If it is rotated along AB it would be a cone

5 0

3 years ago

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What os the correct question and why​

What os the correct question and why