Plz help me thank you. 4 cm of the ice tray

Mathematics

2 answers:

disa [49]3 years ago

6 0

I had to do this and it’s fake, my whole class got it wrong even though we all put the right answer so ya

Readme [11.4K]3 years ago

4 0

I don’t know bud I need point so yea

You might be interested in

A point P is moving along the curve whose equation is y = \sqrt x . Suppose that x is increasing at the rate of 4 units/s when x

Mice21 [21]

Answer:3 units/s

Step-by-step explanation:

Given

$y=\sqrt{x}$

Point P lie on this curve so any general point on curve can be written as $(x,\sqrt{x})$

and $\frac{\mathrm{d} x}{\mathrm{d} t}=4 units/s$

Distance between Point P and (2,0)

$P=\sqrt{(x-2)^2+(\sqrt{x}-0)^2}$

P at x=3 P=2

rate at which distance is changing is

$\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{\mathrm{d} \sqrt{(x-2)^2+(\sqrt{x}-0)^2}}{\mathrm{d} t}$

$\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{2x-3}{\sqrt{(x-2)^2+(\sqrt{x}-0)^2}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{2\times 3-3}{2\times 2}\times 4=3 units/s$

8 0

3 years ago

Algebra 1 CR-8

antiseptic1488 [7]

$\mathrm{Domain\:of\:}\:x^2+2x-15\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

$\mathrm{Range\:of\:}x^2+2x-15:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:-16\:\\ \:\mathrm{Interval\:Notation:}&\:[-16,\:\infty \:)\end{bmatrix}$

$\mathrm{Axis\:interception\:points\:of}\:x^2+2x-15:\quad \mathrm{X\:Intercepts}:\:\left(3,\:0\right),\:\left(-5,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:-15\right)$