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3 years ago

Which of the following does not represent the solution of the compound inequality below?

Mathematics

1 answer:

STALIN [3.7K]3 years ago

8 0

3 - 2x > 17 subtract 3 from both sides

-2x > 14 change the signs

2x < -14 divide both sides by 2

x < -7

5x - 3 > 17 add 3 to both sides

5x > 20 divide both sides by 5

x > 4

Solution: x < -7 or x > 4 → x ∈ (-∞, -7) ∪ (4, ∞) or the graph in the answer A.

<h3>Answer: D.</h3>

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Please help: In PQR, sin P=64, sin R=35, and r=21. Find the length of p.

son4ous [18]

Answer:

32.9 units

Step-by-step explanation:

p/sinP = r/sinR

p = sinP × r/sinR

p = sin(64) × 21/sin(35)

p = 32.9069916

5 0

2 years ago

What is the slope of a line passing through (1,9) and (-3,16) ?

timurjin [86]

Answer:

slope= -7/4

Step-by-step explanation:

y2-y1/x2-x1

16-9/-3-1

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3 years ago

Let f(x) = xe^-x+ ce^-x, where c is a positive constant. For what positive value of c does f have an absolute

Illusion [34]

Answer:

$c=6$

Step-by-step explanation:

The absolute maximum of a continuous function $f(x)$ is where $f'(x)=0$ . Therefore, we must differentiate the function and then set $x=-5$ and $f'(x)=0$ to determine the value of $c$ :

$f(x)=xe^{-x}+ce^{-x}$

$f'(x)=-xe^{-x}+e^{-x}-ce^{-x}$

$0=-(-5)e^{-(-5)}+e^{-(-5)}-ce^{-(-5)}$

$0=5e^{5}+e^{5}-ce^{5}$

$0=e^5(5+1-c)$

$0=6-c$

$c=6$

Therefore, when $c=6$ , the absolute maximum of the function is $x=-5$ .

I've attached a graph to help you visually see this.

7 0

2 years ago

Which expression is equivalent to 8g + 2 + 4g − 2 − 2g?

LekaFEV [45]

Answer:

B) 10g

Step-by-step explanation:

Collect like terms

8g + 4g - 2g = 10g

2 - 2 = 0

10g + 0= 0

4 0

1 year ago

If <img src="https://tex.z-dn.net/?f=%5Cmathrm%20%7By%20%3D%20%28x%20%2B%20%5Csqrt%7B1%2Bx%5E%7B2%7D%7D%29%5E%7Bm%7D%7D" id="Tex

Harman [31]

Answer:

See below for proof.

Step-by-step explanation:

Given:

$y=\left(x+\sqrt{1+x^2}\right)^m$

First derivative

$\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $f(g(x))$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=f'(g(x))\:g'(x)$\\\end{minipage}}$

$\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}$

$\begin{aligned} y_1=\dfrac{\text{d}y}{\text{d}x} & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{2x}{2\sqrt{1+x^2}} \right)\\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{x}{\sqrt{1+x^2}} \right) \\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} \right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(x+\sqrt{1+x^2}\right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m\end{aligned}$

Second derivative

$\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If $y=uv$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}$

$\textsf{Let }u=\dfrac{m}{\sqrt{1+x^2}}$

$\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}$

$\textsf{Let }v=\left(x+\sqrt{1+x^2}\right)^m$

$\implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^m$

$\begin{aligned}y_2=\dfrac{\text{d}^2y}{\text{d}x^2}&=\dfrac{m}{\sqrt{1+x^2}}\cdot\dfrac{m}{\sqrt{1+x^2}}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}\\\\&=\dfrac{m^2}{1+x^2}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\\\\ &=\left(x+\sqrt{1+x^2}\right)^m\left(\dfrac{m^2}{1+x^2}-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\right)\\\\\end{aligned}$

$= \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\right)\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)$

Proof

$(x^2+1)y_2+xy_1-m^2y$

$= (x^2+1) \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m$

$= \left(x+\sqrt{1+x^2}\right)^m\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m$

$= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right]$

$= \left(x+\sqrt{1+x^2}\right)^m\left[0]$

$= 0$

8 0

1 year ago