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2 years ago

What the answer Distance between two points (-5,5) and (3,-3)

Mathematics

2 answers:

vitfil [10]2 years ago

8 0

Answer Is In The Image

Kay [80]2 years ago

5 0

Answer:

Step-by-step explanation:

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Please help if you can!

stealth61 [152]

Given:

The figure of a triangle DEF.

To find:

The coordinate of image of given triangle after reflection over y-axis.

Solution:

The vertices of given triangle are D(1,4), E(4,3), F(2,0).

If a figure is reflected over the y-axis, then the rule of reflection is

$(x,y)\to (-x,y)$

Using this rule, we get

$D(1,4)\to D'(-1,4)$

$E(4,3)\to E'(-4,3)$

$F(2,0)\to F'(-2,0)$

Therefore, the coordinates of image are D'(-1,4), E'(-4,3) and F'(-2,0).

5 0

2 years ago

Find the height of a square pyramid that has a volume of 12 cubic feet and a base length of 3 feet

Andreyy89

I got the height to be 4

8 0

2 years ago

Need help urgently please

kakasveta [241]

Answer:

x = -12

y = 3

Step-by-step explanation:

$1/2x + 6(x+15) = 12\\1/2x + 6x + 90 = 12\\6\frac{1}{2} x + 90 = 12\\6\frac{1}{2} x= -78\\x = -12$

$y = x + 15\\y = -12 + 15\\y = 3$

6 0

1 year ago

Read 2 more answers

Use the definition of the derivative to differentiate v=4/2 pie r^3

nata0808 [166]

I suspect 4/2 should actually be 4/3, since 4/2 = 2, while 4/3 would make V the volume of a sphere with radius r. But I'll stick with what's given:

$\displaystyle \frac{dV}{dr} = \lim_{h\to0} \frac{2\pi(r+h)^3-2\pi r^3}{h}$

$\displaystyle \frac{dV}{dr} = 2\pi \lim_{h\to0} \frac{(r^3+3r^2h+3rh^2+h^3)- r^3}{h}$

$\displaystyle \frac{dV}{dr} = 2\pi \lim_{h\to0} \frac{3r^2h+3rh^2+h^3}{h}$

$\displaystyle \frac{dV}{dr} = 2\pi \lim_{h\to0} (3r^2+3rh+h^2)$

$\displaystyle \frac{dV}{dr} = 2\pi \cdot 3r^2 = \boxed{6\pi r^2}$

In Mathematica, you can check this result via

D[4/2*Pi*r^3, r]

3 0

2 years ago

You are given g(x)=4x^2 + 2x and

Strike441 [17]

Answer:

324

Step-by-step explanation:

Given:

$g(x)=4x^2+2x\\ \\f(x)=\int\limits^x_0 {g(t)} \, dt$

Find:

$f(6)$

First, find f(x):

$f(x)\\ \\=\int\limits^x_0 {g(t)} \, dt\\ \\=\int\limits^x_0 {(4t^2+2t)} \, dt\\ \\=\left(4\cdot \dfrac{t^3}{3}+2\cdot \dfrac{t^2}{2}\right)\big|\limits^x_0\\ \\=\left(\dfrac{4t^3}{3}+t^2\right)\big|\limits^x_0\\ \\= \left(\dfrac{4x^3}{3}+x^2\right)-\left(\dfrac{4\cdot 0^3}{3}+0^2\right)\\ \\=\dfrac{4x^3}{3}+x^2$

Now,

$f(6)\\ \\=\dfrac{4\cdot 6^3}{3}+6^2\\ \\=288+36\\ \\=324$

4 0

2 years ago