(5 1/6-x)*2 7/10-1 3/14=3 2/7

Which of the following demonstrates the identity property?

disa [49]

Answer:

B im pretty sure

Step-by-step explanation:

7 0

2 years ago

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The ideal temperature of a freezer to store a particular brand of ice cream is 0°F, with a fluctuation of no more than 2°F. Whic

yan [13]

Answer:

| x - 0 | ≤ 2

Step-by-step explanation:

Given,

The ideal temperature of the freezer = 0° F,

Also, it can fluctuate by 2° F,

Thus, if x represents the temperature of the freezer,

Then, there can be two cases,

Case 1 : x > 0,

⇒ x - 0 ≤ 2

Case 2 : If x < 0,

⇒ 0 - x ≤ 2,

⇒ -( x - 0 ) ≤ 2,

By combining the inequalities,

We get,

| x - 0 | ≤ 2,

Which is the required inequality.

8 0

3 years ago

Rinni was able to plant 15 flowers in five minutes. At this rate how many flowers because she plant in 45 minutes ANSWER RATS

PSYCHO15rus [73]

Answer:

The correct answer is 135

Step-by-step explanation:

This is because her rate is 15 flowers per 5 minutes, and there is 45 minutes. Therefore, 5 minutes occurs 9 times in 45 minutes and fifteen flowers times 9 groups of five minutes is 135.

3 0

2 years ago

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$1200 deposit at an APR of 4% with quarterly compounding for 2 years.

inessss [21]

Answer:1296

Step-by-step explanation:1200*.04=96

1200+96=1296

8 0

3 years ago

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If x^2y-3x=y^3-3, then at the point (-1,2), (dy/dx)?

zavuch27 [327]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2866883

_______________

   dy
Find —— for an implicit function:
    dx

x²y – 3x = y³ – 3

First, differentiate implicitly both sides with respect to x. Keep in mind that y is not just a variable, but it is also a function of x, so you have to use the chain rule there:

$\mathsf{\dfrac{d}{dx}(x^2 y-3x)=\dfrac{d}{dx}(y^3-3)}\\\\\\
\mathsf{\dfrac{d}{dx}(x^2 y)-3\,\dfrac{d}{dx}(x)=\dfrac{d}{dx}(y^3)-\dfrac{d}{dx}(3)}$

Applying the product rule for the first term at the left-hand side:

$\mathsf{\left[\dfrac{d}{dx}(x^2)\cdot y+x^2\cdot \dfrac{d}{dx}(y)\right]-3\cdot 1=3y^2\cdot \dfrac{dy}{dx}-0}\\\\\\
\mathsf{\left[2x\cdot y+x^2\cdot \dfrac{dy}{dx}\right]-3=3y^2\cdot \dfrac{dy}{dx}}$

   dy
Now, isolate —— in the equation above:
   dx

$\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3=3y^2\cdot \dfrac{dy}{dx}}\\\\\\
\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3-3y^2\cdot \dfrac{dy}{dx}=0}\\\\\\
\mathsf{x^2\cdot \dfrac{dy}{dx}-3y^2\cdot \dfrac{dy}{dx}=-\,2xy+3}\\\\\\
\mathsf{(x^2-3y^2)\cdot \dfrac{dy}{dx}=-\,2xy+3}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{-\,2xy+3}{x^2-3y^2}\qquad\quad for~~x^2-3y^2\ne 0}$

Compute the derivative value at the point (– 1, 2):

x = – 1 and y = 2

$\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{-\,2\cdot (-1)\cdot 2+3}{(-1)^2-3\cdot 2^2}}\\\\\\
\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{4+3}{1-12}}\\\\\\
\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{7}{-11}}\\\\\\\\ \therefore~~\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=-\,\dfrac{7}{11}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: <em>implicit function derivative implicit differentiation chain product rule differential integral calculus</em>

6 0

2 years ago