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2 years ago

Complete each statement using <, >, or =. Use your number line to explain your reasoning

Mathematics

1 answer:

Stels [109]2 years ago

6 0

Answer:

Step-by-step explanation:

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Describe a situation that the expression -15÷(-15) can represent

FrozenT [24]

-15 so that is negative. so that is negative 15÷15 so 15 ÷15 is 1 so it would be -15÷-15=-1

7 0

3 years ago

Consider the following equation. f(x, y) = y3/x, P(1, 2), u = 1 3 2i + 5 j (a) Find the gradient of f. ∇f(x, y) = Correct: Your

BaLLatris [955]

$f(x,y)=\dfrac{y^3}x$

a. The gradient is

$\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath$

$\boxed{\nabla f(x,y)=-\dfrac{y^3}{x^2}\,\vec\imath+\dfrac{3y^2}x\,\vec\jmath}$

b. The gradient at point P(1, 2) is

$\boxed{\nabla f(1,2)=-8\,\vec\imath+12\,\vec\jmath}$

c. The derivative of $f$ at P in the direction of $\vec u$ is

$D_{\vec u}f(1,2)=\nabla f(1,2)\cdot\dfrac{\vec u}{\|\vec u\|}$

It looks like

$\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath$

so that

$\|\vec u\|=\sqrt{\left(\dfrac{13}2\right)^2+5^2}=\dfrac{\sqrt{269}}2$

Then

$D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}$

$\boxed{D_{\vec u}f(1,2)=\dfrac{16}{\sqrt{269}}}$

7 0

3 years ago

What is the simplified value of 5x to the power of 3 over 10x squared

andrey2020 [161]

5x^3/10x^2
*Simplify 5/10 to 1/2
*simplify the exponent so that x^1 is on top
1x/2
FINAL ANSWER:
x/2

4 0

3 years ago

How do you do this? plz I really need help.

Vlada [557]

The top two answers are equivalent, while the bottom one is not. You would use distributive property in the parentheses. If both expressions are the same, they are equivalent

7 0

3 years ago

Which of the following is a root of the polynomial shown below f(x)=x^3+2x^2-x-2

Aleks [24]

F(x) = x^3 + 2x^2 - x - 2
f(x) = x^2(x + 2) - (x + 2) 
f(x) = (x^2 - 1)(x + 2) 
f(x) = (x + 1)(x - 1)(x + 2) 
Roots: -1, 1, -2

3 0

3 years ago

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