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Bad White [126]
2 years ago
15

I’m trying to find value of w

Mathematics
1 answer:
Alex777 [14]2 years ago
5 0

Answer:

w generally indicates a cube root of unity in complex numbers

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PLEASE HELP WORTH 99 POINTS AND CORRECT ANSWER WILL GET BRAINLIEST
Soloha48 [4]

Hello!

I believe the answers are:

A) The vertex is (1, −6)

C) The graph has a minimum.

I hope it helps!

7 0
3 years ago
Read 2 more answers
What set of reflections would carry hexagon ABCDEF onto itself?
k0ka [10]

Answer:

Took FLVS test and got it right

A. y=x, x-axis, y=x, y-axis.

Step-by-step explanation:

Took FLVS test and got it right

A. y=x, x-axis, y=x, y-axis.

8 0
2 years ago
The value of 110 coins, consisting of dimes and quarters, is $20.30. how many of each kind of coin are there?
deff fn [24]
D + q = 110......d = 110 - q
0.10d + 0.25q = 20.30

0.10(110 - q) + 0.25q = 20.30
11 - 0.10q + 0.25q = 20.30
-0.10q + 0.25q = 20.30 - 11
0.15q = 9.30
q = 9.30/0.15
q = 62 <==== there are 62 quarters

d = 110 - q
d = 110 - 62
d = 48 <==== there are 48 dimes

6 0
3 years ago
What is your sister’s total cost under each of the two plans?2. Suppose your sister doubles her monthly usage to 3,500 minutes a
chubhunter [2.5K]

Answer:

(1) The total cost under Plan A is <u>$92</u> and the total cost under Plan B is <u>$515</u>.

(2) The total cost under Plan A is <u>$92</u> and the total cost under Plan B is <u>$1,030</u>.

Step-by-step explanation:

<u><em>The question is incomplete, so the complete question is below:</em></u>

A\ cell \ phone \ company\  offers\  two \ different \ plans.

Plan\ A\ costs\ \$92\ per\ month\ for\ unlimited\ talk\ and\ text.\ Plan\ B\ costs \ $0.20\ per\ minute\ plus\ \$0.10\ per\ text\ message\ sent.You\  need \  to\   purchase \  a \  plan \  for\   your \  14-year-old \  sister.Your\  sister\  currently \ uses\  1,750 \ minutes\  and \ sends\  1,650 \ texts\  each \ month.(1) What\  is\   your \  sister's\   total\   cost\   under \  each\   of\   the\   two\   plans?

(2) Suppose\  your\  sister\  doubles\  her\  monthly\  usage \ to\  3,500\  minutes \ and\  sends\  3,300 \ texts.What \ is \  your  \ sister's \  total \  cost \  under \  each \  of  \ the \  two \  plans?

Now, to find (1) total cost for each of the two plans. (2) Total cost under each of the two plans, if sister doubles her monthly usage to 3,500 minutes and sends 3,300 texts.

<h3>(1) </h3>

As, the Plan A is for unlimited talk and text

Cost of the Plan A = $92.

<u><em>Now, to find the cost under Plan B:</em></u>

According to question:

Rate of call per minute is $0.20 and per text is $0.10.

Calls she uses currently is 1,750 minutes and text 1,650.

<u><em>So, to get the cost of Plan B:</em></u>

0.20\times 1750+0.10\times 1650

=350+165

=515.

Thus, Plan B costs $515.

<h3>(2)</h3>

<u><em>Now, to get the total cost as, sister doubles her monthly usage to 3,500 minutes and sends 3,300 texts.</em></u>

As, the Plan A is same in both cases and is for unlimited text and calls.

So, cost of Plan A = $92.

As, the monthly usage is double.

Calls in minutes are 3,500.

Texts are 3,300.

<u><em>Now, to get the total cost under Plan B:</em></u>

0.20\times 3500+0.10\times 3300

=700+330

=1030.

Hence, the cost of Plan B = $1,030.

Therefore, (1) The total cost under Plan A is $92 and the total cost under Plan B is $515.

(2) The total cost under Plan A is $92 and the total cost under Plan B is $1,030.

7 0
3 years ago
PLEASE SHOW WORK
CaHeK987 [17]

(C)

Step-by-step explanation:

The volume of the conical pile is given by

V = \dfrac{\pi}{3}r^2h

Taking the derivative of V with respect to time, we get

\dfrac{dV}{dt} = \dfrac{\pi}{3}\dfrac{d}{dt}(r^2h)

\:\:\:\:\:\:\:= \dfrac{\pi}{3}\left(2rh\dfrac{dr}{dt} + r^2\dfrac{dh}{dt}\right)

Since r is always equal to h, we can set

\dfrac{dr}{dt} = \dfrac{dh}{dt}

so that our expression for dV/dt becomes

\dfrac{dV}{dt} = \dfrac{\pi}{3}\left(3r^2\dfrac{dh}{dt}\right)

\:\:\:\:\:\:\:= \pi r^2\dfrac{dh}{dt}

Solving for dh/dt, we get

\dfrac{dh}{dt} = \dfrac{1}{\pi r^2}\dfrac{dV}{dt}

\:\:\:\:\:\:\:= \dfrac{1}{9\pi\:\text{m}^2}(36\:\text{m}^3\text{/s})

\:\:\:\:\:\:\:= \dfrac{4}{\pi}\:\text{m/s}

7 0
3 years ago
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