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Mariulka [41]
3 years ago
15

Line segment AB is shown on a coordinate grid:

Mathematics
2 answers:
jenyasd209 [6]3 years ago
7 0

A'B' and AB are equal in length because a rotation does change the length of a line segment it only changes it's position on the graph.

enot [183]3 years ago
5 0

The answer is C.) A'B' and AB are equal in length.

This is true because a rotation doesn't change anything besides the position of the triangle, polygon, or in this case a line. And also I just the quiz and got it correct.

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A=?,b=12mm,h=7.5mm WHAT IS THE ANSWER TO THIS PROBLEM
o-na [289]
0.625?
7.5 ÷ 12 = 0.625
Used a calculator
5 0
3 years ago
Madison planted her garden with 2 5 marigolds and 3 7 tulips. What fraction of the plants in the garden are either marigolds or
vfiekz [6]
To answer this you will combine both fractions together to get a total. The denominators are not the same so you will need to create equivalent fractions for both with a common denominator. The denominator is like the units, and to be able to add them they must be the same.

2/5 + 3/7

2/5 = 14/35 and 3/7 = 15/35

14/35 + 15/35 = 29/35

29/35 of the garden are these 2 flowers.

5 0
3 years ago
Read 2 more answers
Find the point P on the line y=2x that is closest to the point (20,0) .what is the least distance between pans (20,0)?
Vsevolod [243]
The closest point on a line, to another point, will be a point that's on a normal of that line, or a line that is perpendicular to it, notice picture below

so, y = 2x, has a slope of 2, a perpendicular line to it, will have a slope of negative reciprocal that, or \bf 2\qquad negative\implies -2\qquad reciprocal\implies \cfrac{1}{-2}\implies -\cfrac{1}{2}

so, we know that line passes through the point 20,0, and has a slope of -1/2

if we plug that in the point-slope form, we get \bf y-0=-\cfrac{1}{2}(x-20)\implies y=-\cfrac{1}{2}x+10

now, the point that's on 2x and is also on that perpendicular line, is the closest to 20,0 from 2x, thus, is where both graphs intersect, as you can see in the graph

thus  \bf 2x=-\cfrac{1}{2}x+10  solve for "x'

------------------------------------------------------------

not sure on the 2nd part, but sounds like, what's the distance from that point to 20,0, well, if that's the case, just use the distance equation

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ \square }}\quad ,&{{ \square }})\quad 
%  (c,d)
&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}


6 0
3 years ago
A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k
Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0
3 years ago
Round 0.172 to the nearest hundredth.
aleksandr82 [10.1K]
It is <span>Round 0.172 to the nearest hundredth.</span>
8 0
3 years ago
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