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3 years ago

11

Your teacher asked you to find a recipe that includes two ingredients with the ratio of 1/2 cup 1/8 cup give an example of two i

ngredients in a recipe that would meet this requirement

1 answer:

GuDViN [60]3 years ago

4 0

Divide 1/2 by 1/8 by doing Keep change flip. 1/2 divided by 1/8 turns to 1/2 x 8/1 =9/2 and then divide 9/2 like this 9 divided by 2= 2 1/2

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7th grade math, I need some help with this

Mars2501 [29]

Answer:

20%

Step-by-step explanation:

Earlier snow depth = 2 feet

Later snow depth = 1.6 feet

Change

2 - 1.6

0.4 feet

Percentage change

(0.4 / 2) * 100

20%

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3 years ago

In NTUV, the measure of N=90) UV = 8.7 feet, and VT = 3.6 feet. Find the measure of to the nearest degree.

erik [133]

Answer:

5

Step-by-step explanation:

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2 years ago

Read 2 more answers

Which one is greater: 5.098 or 5.4

andreyandreev [35.5K]

Answer:

5.4

Step-by-step explanation:

The number after the decimal is bigger in 5.4 than 5.098 which is the only number that matters in this question however if both numbers were the same you would move to the next number and whichever one has the bigger number.

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2 years ago

Three fourths y plus eight

ludmilkaskok [199]

Answer:

3/4y + 8

Step-by-step explanation: It has no like terms so ye

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3 years ago

Use calculus to find the absolute maximum and minimum values of the function. (round all answers to three decimal places.) f(x)

Allisa [31]

Part A:

Given the function $f(x)=x+2\cos(x)$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow1-2\sin{x}=0 \\ \\ \Rightarrow2\sin{x}=1 \\ \\ \Rightarrow\sin{x}= \frac{1}{2} \\ \\ \Rightarrow x=\sin^{-1}{\frac{1}{2}}= \frac{\pi}{6}$

Using the second derivative test,

$f''(x)=-2cosx \\ \\ \Rightarrow f''\left( \frac{\pi}{6} \right)=-2\cos{\left( \frac{\pi}{6} \right)}=-1.732$

Since the second derivative gives a negative number, the given function has a maximum point at $x=\frac{\pi}{6}$ .

And the maximum point is given by:

$f\left( \frac{\pi}{6} \right)=\frac{\pi}{6}+2\cos\left( \frac{\pi}{6} \right) \\ \\ =0.5236+2(0.8660)=0.5236+1.732 \\ \\ =\bold{2.256}$

i.e. $\left(\frac{\pi}{6},\ 2.256\right)$

Part B:

Given the function $f(x)=e^{-x}-e^{-2x}$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow-e^{-x}+2e^{-2x}=0 \\ \\ \Rightarrow2e^{-2x}=e^{-x} \\ \\ \Rightarrow2e^{-x}=1 \\ \\ \Rightarrow e^{-x}=\frac{1}{2} \\ \\ \Rightarrow-x=\ln \frac{1}{2}=-0.6931 \\ \\ \Rightarrow x=0.6931$

Using the second derivative test,

$f''(x)=e^{-x}-4e^{-2x} \\ \\ \Rightarrow f''(0.6931)=e^{-0.6931}-4e^{-2(0.6931)} \\ \\ =0.5-4e^{-1.386}=0.5-4(0.25)=0.5-1 \\ \\ =-0.5$

Since the second derivative gives a negative number, the given function has a maximum point at $x=0.6931$ .

And the maximum point is given by:

$f(0.6931)=e^{-0.6931}-e^{-2(0.6931)} \\ \\ =0.5-e^{-1.386}=0.5-0.25=\bold{0.25}$

i.e. (0.693, 0.25)

3 0

3 years ago