Given sin r = 0.5, what is cos r?

Subtract the following. Write your answer in standard notation . 18-9.2×10^-4

puteri [66]

Answer:1.799908×10^1

Step-by-step explanation:

18-0.00092

=17.99908

In standard form=1.799908×10^1

8 0

2 years ago

Neelam's daily income is ₹ 800. Her daily savings amount is ₹100. What is the ratio of her SAVINGS to her EXPENDITURE?

sveta [45]

Answer: ratio of her SAVINGS to her EXPENDITURE=1:7

Step-by-step explanation:

Neelam's daily income = ₹ 800

Neelam's daily savings amount = ₹100

Therefore her expenditure will be Daily income - Daily savings

= ₹ 800- ₹ 100= ₹ 700

ratio of her SAVINGS to her EXPENDITURE

= 100: 700

=1:7

3 0

2 years ago

The length of a rectangle is 3 meters greater than 2 times the width. The perimeter of the rectangle is 30 meters.What is the le

faust18 [17]

Answer:

11 meters

Step-by-step explanation:

Lets say that w = width of the rectangle, to start. If the length of the rectangle is 3 meters greater than 2 times the width, the length of the rectangle is equal to 3 + 2w.

The perimeter of the rectangle is 2 * length of rectangle + 2 * width of the rectangle. With the perimeter being equal to 30 and width being w and length being 2w+3:

The perimeter of the rectangle is 2(w) + 2(2w+3) = 30.

We first need to find out w first, which will give us the width of the rectangle. Taking it step by step, we get:

2w + 4w + 6 = 30

6w + 6= 30

6w = 24 which is done by subtracting both sides by 6 to put the variables on one side and the values on the other side

w = 4 which is done by dividing 6 on both sides

Ultimately, this gets width to be 4 meters. Now that we found the width, we need to plug w = 4 into the equation we set up for length which is 2w+3.

That being said, the ANSWER is:

length of rectangle = 2(4)+3 = 11 meters

Hope this helps! :)

4 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 $(2,10,\pm2\sqrt5)$ .