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3 years ago

Write a multiplication sentence that shows the zero property of multiplication. Explain why it shows this property.

1 answer:

4 0

Okay so the zero property of multiplication is anything times zero equals zero, so just use a number, multiply it times zero, and show what it equals (which is zero) explain that it shows the property because the variable is multiplied times zero.

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Which shows the equation y – x = 3 written in slope-intercept form? Question 1 options: y – x – 3 = 0 y – 2 = x + 3 y = 3x y = x

Kazeer [188]

Y=x+3
you need to add the x to the opposite side to get y by itself

4 0

3 years ago

What's one third multiply 3

allsm [11]

Simple! 1/3 times 3 equals 3/3, or 1.

7 0

3 years ago

Read 2 more answers

Solve for x in the diagram below

ella [17]

Answer:

Step-by-step explanation:

3x + x + 100 = 180 (sum of angles on a straight line)

4x = 180-100

4x=80

divide both sides by 4

x=20

8 0

3 years ago

Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)

leonid [27]

Let $P=(x,y,z)$ be an arbitrary point on the surface. The distance between $P$ and the given point $(7,11,0)$ is given by the function

$d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}$

Note that $f(x)$ and $f(x)^2$ attain their extrema, if they have any, at the same values of $x$ . This allows us to consider the modified distance function,

$d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2$

So now you're minimizing $d^*(x,y,z)$ subject to the constraint $z^2=xy$ . This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

$\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)$

which has partial derivatives

$\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}$

Setting all four equation equal to 0, you find from the third equation that either $z=0$ or $\lambda=-1$ . In the first case, you arrive at a possible critical point of $(0,0,0)$ . In the second, plugging $\lambda=-1$ into the first two equations gives

$\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10$

and plugging these into the last equation gives

$z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5$

So you have three potential points to check: $(0,0,0)$ , $(2,10,2\sqrt5)$ , and $(2,10,-2\sqrt5)$ . Evaluating either distance function (I use $d^*$ ), you find that

$d^*(0,0,0)=170$
$d^*(2,10,2\sqrt5)=46$
$d^*(2,10,-2\sqrt5)=46$

So the two points on the surface $z^2=xy$ closest to the point $(7,11,0)$ are $(2,10,\pm2\sqrt5)$ .

5 0

4 years ago

Let U = {a, b, c, e, d, e, f, g, h, i, j, k}; A = {a, b, e, f, h, j}; B = {b, c, d, f, j}. Draw a Venn Diagram with a rectangle

marishachu [46]

<h3>Answer: Venn Diagram is below</h3>

The rectangle represents the universal set (marked as "set U"). Everything is contained in this rectangle.

Circle A represents set A = {a, b, e, f, h, j}

Circle B represents set B = {b, c, d, f, j}

The common elements between the two sets are {b, f, j}, so we'll write these three values in the overlapping portion of the two circles.

The items {a,e,h} will go in circle A, but not in the overlapping portion since they are found only in set A.

Similarly we'll have {c,d} go in circle B but not in the overlapping portion since these items are only found in set B.

The remaining terms {g, i, k} go outside both circles, but inside the rectangle, because neither of these elements are in set A or set B.

6 0

3 years ago