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3 years ago

160% of 1825 rounded to nearest tenth

2 answers:

8 0

I got the answer 2,920. did that work? because you have to take 1825 times 1.6 which is also taking 160 percent of 1825 and that is the answer i got. I hope it helps dear!!!

lutik1710 [3]3 years ago

4 0

I simply put 160% x 1825 into a calculator and got 2920

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One canned juice drink is 25% orange juice; another is 10% orange juice. How many liters of each should be mixed together in ord

Vikki [24]

Let's say we use "x" of the 25% OJ... ok... well, how much juice is really in that "x" liters? well, 25% of it is OJ, the rest is water, how much is 25% of x? well, (25/100) * x, or 0.25x.

likewise, we'll use "y" of the 10% OJ, and it will end up with (10/100) *y or 0.10y of juice in it.

whatever "x" and "y" are, they must add up to 15 Liters, and the juice concentration on that mixture, must also be the sum of their sum concentration.

$\bf \begin{array}{lccclll}
&\stackrel{liters}{amount}&\stackrel{juice~\%}{quantity}&\stackrel{juice}{quantity}\\
&------&------&------\\
\textit{25\% juice}&x&0.25&0.25x\\
\textit{10\% juice}&y&0.10&0.10y\\
------&------&------&------\\
mixture&15&0.17&2.55
\end{array}
\\\\\\
\begin{cases}
x+y=15\implies \boxed{y}=15-x\\
0.25x+0.10y=2.55\\
----------\\
0.25x+0.10\left(\boxed{15-x} \right)=2.55
\end{cases}
\\\\\\
0.25x-0.10x+1.5=2.55\implies 0.15x=1.05\implies x=\cfrac{1.05}{0.15}
\\\\\\
x=7$

how much will be it of the 10% juice? well, y = 15 - x.

6 0

2 years ago

I apologize for all the questions I asked

Paul [167]

Answer:

Step-by-step explanation:

It's not clear but yeah it's fine if you ask a lot cause everytime I see this I know it's you :)

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3 years ago

A parallelogram has symmetry with respect to the point of intersection of its diagonals. t/f?

Anna007 [38]

It is true that a parallelogram has symmetry with respect to the point of intersection of its diagonals.

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2 years ago

Kate took out a subsidized Stafford loan worth $9,710 to pay for college. The interest rate on the loan was 5.9%, compounded mo

NikAS [45]

<span>If Kate took out a loan of $9,710 and it took her 5 years to pay off she would have to pay 60 months of interest. Given that, the interest rate of 5.9% would of been $572.89 the first month. Use this information to find the answer.

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</span>

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2 years ago

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

5 0

2 years ago