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2 years ago

Pa help po plss help me plss plsss help me

SAT

2 answers:

lukranit [14]2 years ago

7 0

Answer:

3 is actually b, because its translucent, yet not transparent

alukav5142 [94]2 years ago

4 0

1) B.Metallic windows
2) D. Absorption
3) A. Clear glass

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What it means that [If x is a factor of p(x), what is the value of k?]

kompoz [17]

Answer:

$k = -4$

Explanation:

Given

$P(x) = (3x^2 - 5)(x + k) - 20$

x is a factor

Required

Solve for k

x is a factor means that the polynomial can be divided successfully by x.

In other words, x is a zero of the polynomial.

i.e.

$x = 0$ and $P(x) = 0$

Substitute 0 for x in P(x)

$P(x) = (3x^2 - 5)(x + k) - 20$

$P(x) = (3*0^2 - 5)(0 + k) - 20$

$P(x) = (0- 5)(k) - 20$

$P(x) = (-5)(k) - 20$

$P(x) = -5k - 20$

Substitute 0 for P(x)

$0 = -5k - 20$

Collect Like Terms

$5k = - 20$

Divide both sides by 5

$k = -4$

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Answer:

in blood

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i don't know but just trying

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2 years ago

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3 years ago

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The point (−2,4) lies on the curve in the xy-plane given by the equation f(x)g(y)=17−x−y, where f is a differentiable function o

Nikitich [7]

The value of $\frac{dy}{dx}$ is -3. $\blacksquare$

<h2>Procedure - Differentiability</h2><h3 /><h3>Chain rule and derivatives</h3><h3 />

We derive an expression for $\frac{dy}{dx}$ by means of chain rule and differentiation rule for a product of functions:

$\frac{d}{dx}[f(x)\cdot g(y)] = [f'(x)\cdot \frac{dx}{dx}]\cdot g(y) + f(x) \cdot [g'(y)\cdot \frac{dy}{dx} ]$

$\frac{d}{dx}[f(x)\cdot g(y)] = f'(x)\cdot g(y) +f(x)\cdot g'(y)\cdot \frac{dy}{dx}$ (1)

If we know that $f(x) \cdot g(y) = 17-x-y$ , $f(-2) = 3$ ,<em> </em> $f'(-2) = 4$ ,<em> </em> $g(4) = 5$ and $g'(4) = 2$ , then we have the following expression:

$-1-\frac{dy}{dx} = (4)\cdot (5) + (3)\cdot (2) \cdot \frac{dy}{dx}$

$-1-\frac{dy}{dx} = 20 + 6\cdot \frac{dy}{dx}$

$7\cdot \frac{dy}{dx} = -21$

$\frac{dy}{dx} = -3$

The value of $\frac{dy}{dx}$ is -3. $\blacksquare$

To learn more on differentiability, we kindly invite to check this verified question: brainly.com/question/24062595

<h3>Remark</h3>

The statement is incomplete and full of mistakes. Complete and corrected form is presented below:

<em>The point (-2, 4) lies on the curve in the xy-plane given by the equation </em> $f(x)\cdot g(y) = 17 - x\cdot y$ <em>, where </em> $f$ <em> is a differentiable function of </em> $x$ <em> and </em> $g$ <em> is a differentiable function of </em> $y$ <em>. Selected values of </em> $f$ <em>, </em> $f'$ <em>, </em> $g$ <em> and </em> $g'$ <em> are given below: </em> $f(-2) = 3$ <em>, </em> $f'(-2) = 4$ <em>, </em> $g(4) = 5$ <em>, </em> $g'(4) = 2$ <em>. </em>