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lions [1.4K]
3 years ago
9

Solve the formula for the indicated variable  c=yt+y,for t (the solution is t=___)

Mathematics
1 answer:
Vlad1618 [11]3 years ago
7 0
C=yt+y. Divide both sides by y and u get c/y=t+1 subtract 1. The ANSWER IS c/y-1=t
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Converges and diverged part 4<br>​
Allushta [10]

Answer:

Diverges; no sum

Step-by-step explanation:

This is comparable to:

\sum_{k=1}^\infty a \cdot r^{k-1} where:

r is the common ratio and a is the first term.

The series converges to:

\text{First term}\cdot \frac{1}{1-\text{common ratio}}

if the ratio's absolute value is less than 1.

This is a geometric series.

The common ration is -1.04 .

The first term in the series is 0.001.

Since the absolute value of -1.04 is 1.04>1, the series diverges.

4 0
3 years ago
1. A man purchased a TV set at Rs 4000 and sold it for Rs 2500. Find his loss per cent.
DochEvi [55]

Answer:

1: 37.5%

2: C.p= 12×25= 300

  S.p= 12×28= 336

profit%=336-300×100/300=12% ansr..

3: Therefore, Mr. Arsalan sold the plot at Rs 129600.

Step-by-step explanation:

BRO HOPE IT HELPED I COULDN'T WRITE ON PAPER, SORRY FOR THAT...

5 0
3 years ago
Will someone please help meeeeeee
Anon25 [30]

Answer:

x = 5

Step-by-step explanation:

x/2 = 15/6

6 * x = 15 * 2

6x = 30

x = 30 / 6

x = 5

6 0
2 years ago
Read 2 more answers
Can (somebody)help me!!!
Mariulka [41]

Answer:

20 meters

Step-by-step explanation:

The area of a rhombus is the height multiplied by the base. Conversely, the height of a rhombus is the area divided by the base. Therefore, the height of this rhombus is 20 meters. Hope this helps!

8 0
3 years ago
a dump truck weighs 11.25 tons. a conveyor belt pours sand into the truck at a constant rate of 1/4 ton per minute. the dump tru
Serhud [2]
We know these variables:

W1B = Weight of the truck 1 before starting the filling.
W1A = Weight of the truck 1 when filled.
R1F =  Rate of change (Truck 1 being filled)

W2B = Weight of the truck 2 before starting the filling.
W2A = Weight of the truck 2 when filled.
R2F =  Rate of change (Truck 2 being filled)

Given that a conveyor belt pours sand into the truck and stops when filled, then the amount of sand, when the truck is filled to capacity, is<span>:

</span>S =  W_{1B} - W_{1A} = 18-11.25 = 6.75Tons
<span>
Let's analyze Truck 1:

Given that the rate of change is:

</span>\frac{1}{4}  ton/min 
<span>
This means that in 1min the amount of sound poured is 1/4Ton, then the time </span>elapsed when filled using rule of three is:

t =  \frac{6.75}{ \frac{1}{4} } = 27min

A similar analysis happens with Truck 2: 

Given that the rate of change is:

\frac{1}{8} ton/min 

Then:

t = \frac{6.75}{ \frac{1}{8}} = 54min

Now we can find two straight lines:

Truck 1:

\left \{ {{t=0min} \atop {S=0Ton}} \right.

\left \{ {{t=27min} \atop {S=6,75Ton} \right

S-S_{0} = m(t-t_{0})

S-0 = \frac{6.75-0}{27-0} (t-0)

∴ S = 0.25t

Truck 2:

\left \{ {{t=0min} \atop {S=6.75Ton}} \right

\left \{ {{t=54min} \atop {S=0Ton}} \right

S-S_{0} = m(t-t_{0})

S-6.75 = \frac{6.75-0}{0-54} (t-0)

∴ S = -0.125t+6.75

<span>The trucks are the same weight when they have the same amount of sand, so solving for t:
</span>
0.25t = -0.125t+6.75
0.375t = 6.75
t = 18min

Finally, the amount of sand in each truck:

S = 0.25(18) = 4.5Ton
3 0
3 years ago
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