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3 years ago

if the equation 2x^2+ bx+ 5=0 has no real solutions, which of the following must be true A. b^2 < 10 B. b^2> 10 C. b^240

Mathematics

1 answer:

melisa1 [442]3 years ago

6 0

Answer:

C. b²< 40

Step-by-step explanation:

2x²+ bx + 5=0 has no real solutions

=> D< 0

b² < 4ac

b²< 4(2)(5)

b²< 40

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HELP MEeeeeeeeee g: R² → R a differentiable function at (0, 0), with g (x, y) = 0 only at the point (x, y) = (0, 0). Consider<im

GrogVix [38]

(a) This follows from the definition for the partial derivative, with the help of some limit properties and a well-known limit.

• Recall that for $f:\mathbb R^2\to\mathbb R$ , we have the partial derivative with respect to $x$ defined as

$\displaystyle \frac{\partial f}{\partial x} = \lim_{h\to0}\frac{f(x+h,y) - f(x,y)}h$

The derivative at (0, 0) is then

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(0+h,0) - f(0,0)}h$

• By definition of $f$ , $f(0,0)=0$ , so

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(h,0)}h = \lim_{h\to0}\frac{\tan^2(g(h,0))}{h\cdot g(h,0)}$

• Expanding the tangent in terms of sine and cosine gives

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{h\cdot g(h,0) \cdot \cos^2(g(h,0))}$

• Introduce a factor of $g(h,0)$ in the numerator, then distribute the limit over the resulting product as

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{g(h,0)^2} \cdot \lim_{h\to0}\frac1{\cos^2(g(h,0))} \cdot \lim_{h\to0}\frac{g(h,0)}h$

• The first limit is 1; recall that for $a\neq0$ , we have

$\displaystyle\lim_{x\to0}\frac{\sin(ax)}{ax}=1$

The second limit is also 1, which should be obvious.

• In the remaining limit, we end up with

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)}h = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h$

and this is exactly the partial derivative of $g$ with respect to $x$ .

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h = \frac{\partial g}{\partial x}(0,0)$

For the same reasons shown above,

$\displaystyle \frac{\partial f}{\partial y}(0,0) = \frac{\partial g}{\partial y}(0,0)$

(b) To show that $f$ is differentiable at (0, 0), we first need to show that $f$ is continuous.

• By definition of continuity, we need to show that

$\left|f(x,y)-f(0,0)\right|$

is very small, and that as we move the point $(x,y)$ closer to the origin, $f(x,y)$ converges to $f(0,0)$ .

We have

$\left|f(x,y)-f(0,0)\right| = \left|\dfrac{\tan^2(g(x,y))}{g(x,y)}\right| \\\\ = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)^2}\cdot\dfrac{g(x,y)}{\cos^2(g(x,y))}\right| \\\\ = \left|\dfrac{\sin(g(x,y))}{g(x,y)}\right|^2 \cdot \dfrac{|g(x,y)|}{\cos^2(x,y)}$

The first expression in the product is bounded above by 1, since $|\sin(x)|\le|x|$ for all $x$ . Then as $(x,y)$ approaches the origin,

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{|g(x,y)|}{\cos^2(x,y)} = 0$

So, $f$ is continuous at the origin.

• Now that we have continuity established, we need to show that the derivative exists at (0, 0), which amounts to showing that the rate at which $f(x,y)$ changes as we move the point $(x,y)$ closer to the origin, given by

$\left|\dfrac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}\right|,$

approaches 0.

Just like before,

$\left|\dfrac{\tan^2(g(x,y))}{g(x,y)\sqrt{x^2+y^2}}\right| = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)}\right|^2 \cdot \left|\dfrac{g(x,y)}{\cos^2(g(x,y))\sqrt{x^2+y^2}}\right| \\\\ \le \dfrac{|g(x,y)|}{\cos^2(g(x,y))\sqrt{x^2+y^2}}$

and this converges to $g(0,0)=0$ , since differentiability of $g$ means

$\displaystyle \lim_{(x,y)\to(0,0)}\frac{g(x,y)-g(0,0)}{\sqrt{x^2+y^2}}=0$

So, $f$ is differentiable at (0, 0).

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3 years ago

3√12•√6<br><br>please show a step by step explanation

Oduvanchick [21]

Answer: $18\sqrt{2}\\\\$

This is the same as saying 18*sqrt(2) or you could say 18√2

============================================================

Explanation:

We'll be using this square root rule: $\sqrt{A}*\sqrt{B} = \sqrt{A*B}$

So we could have this as our steps:

$3\sqrt{12}*\sqrt{6}\\\\3\sqrt{12*6} \ \text{ ... Apply the rule mentioned}\\\\3\sqrt{72}\\\\3\sqrt{36*2}\\\\3\sqrt{36}*\sqrt{2} \ \text{ ... Apply the rule again, but in reverse}\\\\3*6*\sqrt{2}\\\\18\sqrt{2}\\\\$

Therefore, $3\sqrt{12}*\sqrt{6} = 18\sqrt{2}$

The basic idea is to first combine the roots using that rule. From there, we factor the radicand such that we pull out the largest perfect square factor. This will allow us to break the root apart and fully simplify it.

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Express the following linear equations in the form ax+by+c= 0 and indicate the values of a, b and c in each case: (1) y = x/5

EleoNora [17]

Answer:

Did the question get cut off?

a = -(1/5)

b = 1

c = 0

Step-by-step explanation:

y = x/5 may be rewritten as y = (1/5)x

y = (1/5)x

y - (1/5)x = 0

- (1/5)x + y + 0 = 0

ax + by + c = 0

a = -(1/5)

b = 1

c = 0

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mihalych1998 [28]

Answer:

Step-by-step explanation:

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if the equation 2x^2+ bx+ 5=0 has no real solutions, which of the following must be true A. b^2 < 10 B. b^2> 10 C. b^240​

if the equation 2x^2+ bx+ 5=0 has no real solutions, which of the following must be true A. b^2 < 10 B. b^2> 10 C. b^240