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2 years ago

What is the distance between (22,-3) and (1, -3)?

Mathematics

1 answer:

Bingel [31]2 years ago

6 0

Answer:

Option D

Step-by-step explanation:

<h2>Distance between two points</h2>

The y-value of the two points is unchanged. So, the line is parralel to x-axis.

The distance will be the diffrence between the x-co ordinates.

$\sf Distance = 1 - (-2\dfrac{1}{2})$

$\sf =1+\dfrac{5}{2}\\\\= \dfrac{2}{2}+\dfrac{5}{2}\\\\=\dfrac{7}{2}\\\\=3\dfrac{1}{2}$

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(1.7 x 10^4) x (8.5 x 10 ^2)

vlabodo [156]

Answer:

144500

Step-by-step explanation:

8 0

3 years ago

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May anyone help me on this one

ivanzaharov [21]

The answer is 1 1/2 and 1 whole

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3 years ago

A nickel has a mass of 5 grams.What is the mass of a nickel in milligrams?

DiKsa [7]

5000........................

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4 years ago

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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

I never got my anwser to finding the product and write the lowest term to<br> 7/54×27/35

kenny6666 [7]

Answer:

Step-by-step explanation:

7 / 54 * 27/35 =

1/2 * 1/5 =

1/10 <===

4 0

4 years ago