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stiv31 [10]
2 years ago
5

PLEASE HELP ONE QUESTION!!!!!! PLEASEEEEE

Mathematics
2 answers:
In-s [12.5K]2 years ago
7 0
VW=CD since they are congruent figure, so CD=6
Olegator [25]2 years ago
5 0

Answer: 6 units

Step-by-step explanation:

Given : Figure TUVW ≅ figure ABCD,

AB = 12 units , CD = 6 units , and DA = 4 units .

We know that if two polygons are congruent then their corresponding parts are also congruent.

Therefore, if Figure TUVW ≅ figure ABCD, then its corresponding sides must be congruent.

Since the corresponding sides of VW is CD in figure ABCD.

Then the measure of VW = CD = 6 units

Hence, the measure of VW= 6 units

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At a local amusement park, 70% of the rides are made for kids who are 48 inches and taller. If there are 80 rides in total, how
Musya8 [376]

Answer:

56

Step-by-step explanation:

Basically you want to find out what is 70% of 80.

first you have to set up your problem

\frac{n}{80} = \frac{70}{100}

the first fraction would be the percent (70) over 100. (you always use 100)

The fraction to the left would be n over 80. the denominator (the number on the bottom) would be the whole amount. n would be the unknown number

To solve you would use the butterfly method. 80 x 70 and 100 x n. 80 x 70 is 5600 and 100 x n is 100n.

The final step is dividing. 5600 divided by 100 giving you 56

8 0
2 years ago
To rent a car for a trip, four friends are combining their money. The group chat shows the amount of money that each puts in. On
abruzzese [7]

Answer:

1. 189+224+87  2.  500- 189-224

B. q=87

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Silver's gym charges a $65 sign-up fee and $50 per month for their membership. Solar Fitness charges a $20 sign up fee and $55 p
alexandr1967 [171]
65+50x = 55x + 20
-50x -50
65=5x+20
- 20 - 20
55=5x
55/5= 5x/5
11=x
The solution represents that it 11 months will make both gyms the exact same price.
7 0
3 years ago
Combine like terms 4a2-3b2+ 4b + 10a2
jolli1 [7]

Answer:

14a² - 3b² + 4b

Step-by-step explanation:

Like terms are the terms that have similar stuff on it, like same variables, no variables, and stuff like that.

4a2-3b2+ 4b + 10a2

4a² and 10a² is similar. 4a² + 10a² = 14a²

Now, the equation is 14a² - 3b² + 4b. There is no more like terms left, so that is the answer.

Hope this helps!!

-Ketifa

4 0
2 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
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