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3 years ago

Y varies directly with x, and y = 7 when x = 2. What is the value of y when x = 5?

1 answer:

3 0

Y varies directly with x" means y = constant * x or y = kx ----- (1)1. when x = -1/2, y = 2Put it in (1):2 = k * (-1/2)multiply by 2:4 = k * (-1)4 = -kk = -4y = -4x
"find the value of y when x= -0.3"y = -4x = (-4) * (-0.3) = 1.2

So for each problem, start with y = kxThey give you one x and one y value.Put it in y = kx and find k.Then put x = -0.3 and find y.

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Which trend is the plot

xz_007 [3.2K]

Answer:

positive trend

Step-by-step explanation:

the points are all positive

4 0

3 years ago

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Can anyone help me solve this.I need proper equation..Thanks

artcher [175]

Answer:

y = 9 $\sqrt{2}$

Step-by-step explanation:

Using the sine ratio in the right triangle and the exact value

sin45° = $\frac{1}{\sqrt{2} }$

sin45° = $\frac{opposite}{hypotenuse}$ = $\frac{9}{y}$

Multiply both sides by y

y × sin45° = 9, that is

y × $\frac{1}{\sqrt{2} }$ = 9

Multiply both sides by $\sqrt{2}$

y = 9 $\sqrt{2}$

3 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>

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3 years ago

What is 0.045 in scientific notation

strojnjashka [21]

4.5 x 10 exponent -2

8 0

3 years ago

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How do you work this 68.32/2.8=

timurjin [86]

It will be 24.4
I hope this helps

4 0

3 years ago