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1 year ago

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A recipe calls for 5 ounces of vinegar and 10 ounces of oil. Daren is making dressing with 9 ounces of vinegar using the same ra

tio of oil to vinegar.

1 answer:

g100num [7]1 year ago

6 0

Answer:

Step-by-step explanation:

Set this up as ratio of vinegar to oil in fraction form:

$\frac{v}{o}:\frac{5}{10}$

That's what we're given. If we are looking to find how much oil he needs if he's using 9 ounces of vinegar, then 9 goes on top with the vinegar stuff and x goes on bottom as the unknown amount of oil:

$\frac{v}{o}:\frac{5}{10}=\frac{9}{x}$

Cross multiply to get

5x = 90 and

x = 18

Which you probably could do without the proportions. If he is using 5 ounces of vinegar and double that amount of oil, then it just makes sense that if he uses 9 ounces of vinegar he will double that amount in oil to use 18 ounces.

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Use green's theorem to compute the area inside the ellipse x252+y2172=1. use the fact that the area can be written as ∬ddxdy=12∫

Pavel [41]

The area of the ellipse $E$ is given by

$\displaystyle\iint_E\mathrm dA=\iint_E\mathrm dx\,\mathrm dy$

To use Green's theorem, which says

$\displaystyle\int_{\partial E}L\,\mathrm dx+M\,\mathrm dy=\iint_E\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

( $\partial E$ denotes the boundary of $E$ ), we want to find $M(x,y)$ and $L(x,y)$ such that

$\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

and then we would simply compute the line integral. As the hint suggests, we can pick

$\begin{cases}M(x,y)=\dfrac x2\\\\L(x,y)=-\dfrac y2\end{cases}\implies\begin{cases}\dfrac{\partial M}{\partial x}=\dfrac12\\\\\dfrac{\partial L}{\partial y}=-\dfrac12\end{cases}\implies\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

The line integral is then

$\displaystyle\frac12\int_{\partial E}-y\,\mathrm dx+x\,\mathrm dy$

We parameterize the boundary by

$\begin{cases}x(t)=5\cos t\\y(t)=17\sin t\end{cases}$

with $0\le t\le2\pi$ . Then the integral is

$\displaystyle\frac12\int_0^{2\pi}(-17\sin t(-5\sin t)+5\cos t(17\cos t))\,\mathrm dt$

$=\displaystyle\frac{85}2\int_0^{2\pi}\sin^2t+\cos^2t\,\mathrm dt=\frac{85}2\int_0^{2\pi}\mathrm dt=85\pi$

###

Notice that $x^{2/3}+y^{2/3}=4^{2/3}$ kind of resembles the equation for a circle with radius 4, $x^2+y^2=4^2$ . We can change coordinates to what you might call "pseudo-polar":

$\begin{cases}x(t)=4\cos^3t\\y(t)=4\sin^3t\end{cases}$

which gives

$x(t)^{2/3}+y(t)^{2/3}=(4\cos^3t)^{2/3}+(4\sin^3t)^{2/3}=4^{2/3}(\cos^2t+\sin^2t)=4^{2/3}$

as needed. Then with $0\le t\le2\pi$ , we compute the area via Green's theorem using the same setup as before:

$\displaystyle\iint_E\mathrm dx\,\mathrm dy=\frac12\int_0^{2\pi}(-4\sin^3t(12\cos^2t(-\sin t))+4\cos^3t(12\sin^2t\cos t))\,\mathrm dt$

$=\displaystyle24\int_0^{2\pi}(\sin^4t\cos^2t+\cos^4t\sin^2t)\,\mathrm dt$

$=\displaystyle24\int_0^{2\pi}\sin^2t\cos^2t\,\mathrm dt$

$=\displaystyle6\int_0^{2\pi}(1-\cos2t)(1+\cos2t)\,\mathrm dt$

$=\displaystyle6\int_0^{2\pi}(1-\cos^22t)\,\mathrm dt$

$=\displaystyle3\int_0^{2\pi}(1-\cos4t)\,\mathrm dt=6\pi$

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1 year ago

Find the midpoint between each pair of points.<br> (-1,-5) and (-5,9)

Paul [167]

Answer:

work is pictured and shown

4 0

1 year ago

A botanist collected one leaf at random from each of 10 randomly selected mature maple trees of the same species. The mean and t

Zarrin [17]

Answer:

One sample t-test for population mean would be the most appropriate method.

Step-by-step explanation:

Following is the data which botanist collected and can use:

Sample mean
Sample Standard Deviation
Sample size (Which is 10)
Distribution is normal

We have to find the best approach to construct the confidence interval for one-sample population mean. Two tests are used for constructing the confidence interval for one-sample population mean. These are:

One-sample z test for population mean
One-sample t test for population mean

One sample z test is used when the distribution is normal and the population standard deviation is known to us. One sample t test is used when the distribution is normal, population standard deviation is unknown and sample standard deviation is known.

Considering the data botanist collected, One-sample t test would be the most appropriate method as we have all the required data for this test. Using any other test will result in flawed intervals and hence flawed conclusions.

Therefore, One-sample t-test for population mean would be the most appropriate method.

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2 years ago

Complete the statements for the graph of f(x) = |x|. The domain of the function is . The range of the function is . The graph is

ioda

Answer:

Domain of f(x): All real numbers

Range of f(x): $f(x) \geq0$ ---> [0, ∞)

The graph is over the interval (0, ∞)

Step-by-step explanation:

The function absolute value of x that is written: f(x) = | x | assigns only positive values to f(x) at any value of x.

For example, if for f(x) = | x | we make

f(-23) = | -23 | then the absolute value of x "transforms" the number -23 to +23.

Then f(-23) = | -23 | = 23

This means that f(x) will be positive for any value of x.

Therefore, we can conclude that:

Domain of f(x): All real numbers

Range of f(x): $f(x) \geq0$ ---> [0, ∞)

The graph is over the interval (0, ∞)

5 0

1 year ago

Read 2 more answers

At what rate would you need to invest $12000 and make $2880 after 8 years?

lora16 [44]

Answer:

$3\%$

Step-by-step explanation:

we know that

The simple interest formula is equal to

$I=P(rt)$

where

I is the Final Interest Value

P is the Principal amount of money to be invested

r is the rate of interest

t is Number of Time Periods

in this problem we have

$t=8\ years\\ P=\$12,000\\I=\$2,880\\r=?$

substitute in the formula above

$2,880=12,000(8r)$

Solve for t

$2,880=96,000(r)$

$r=2,880/96,000$

$r=0.03$

Convert to percent form

$0.03*100=3\%$

4 0

2 years ago