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3 years ago

The last answer I got for this was not help full. Please help. Will give brainliest.

Mathematics

1 answer:

dimulka [17.4K]3 years ago

7 0

Suppose JJ = Jhalil , J = Joman

1 - JJ 860 J620

2 - JJ 870 J 670

3- JJ880 J 720

4- JJ890 J 770

5- JJ900 J 820

6- JJ910 J 870

7- JJ920 J 920 - Catched up

8- JJ 930 J 970

9- JJ940 J 1020

10- JJ 950 J 1070

11- JJ960 J 1120

12- JJ 970 J 1170

so a - week 7

b- Joman

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Eli has saved $48.50. Eli's brother borrowed 1/10 of Eli's savings. How much money does Eli have left?

zimovet [89]

Answer:

$43.65

Step-by-step explanation:

1. divide 48.50 by 10

4.85

2. subtract 4.85 from 48.50

3. your answer would be $43.65

5 0

3 years ago

Read 2 more answers

How much of a radioactive kind of thorium will be left after 14,680 years if you start with

babymother [125]

Answer:

8978 grams

Step-by-step explanation:

The equation to find the half-life is:

$N(t)= N_{0}e^{-kt}$

N(t) = amount after the time t

$N_{0}$ = initial amount of substance

t = time

It is known that after a half-life there will be twice less of a substance than what it intially was. So, we can get a simplified equation that looks like this, in terms of half-lives.

$N(t)= N_{0}e^{-\frac{ln(\frac{1}{2}) }{t_{h} } t}$ or more simply $N(t)= N_{0}(\frac{1}{2})^{\frac{1}{t_{h} } }$

$t_{h}$ = time of the half-life

We know that $N_{0}$ = 35,912, t = 14,680, and $t_{h}$ =7,340

Plug these into the equation:

$N(t) = 35912(\frac{1}{2})^{\frac{14680}{7340} }$

Using a calculator we get:

N(t) = 8978

Therefore, after 14,680 years 8,978 grams of thorium will be left.

Hope this helps!! Ask questions if you need!!

8 0

2 years ago

A culture started with 2,000 bacteria. After 8 hours, it grew to 2,400 bacteria. Predict how many bacteria will be present after

Debora [2.8K]

Answer:

2950 [Starting from 2,000 bacteria.]

Step-by-step explanation:

400/8 = 50

Bacteria grown/time = bacteria per hour

19x50 = 950

time x bacteria per hour = bacteria grown

2000 + 950

bacteria started + bacteria grown

6 0

3 years ago

Given T5 = 96 and T8 = 768 of a geometric progression. Find the first term,a and the common ratio,r.

Alona [7]

Answer:

Of the given geometric sequence, the first term a is 6 and its common ratio r is 2.

Step-by-step explanation:

Recall that the direct formula of a geometric sequence is given by:

$\displaystyle T_ n = ar^{n-1}$

Where Tₙ is the nth term, a is the initial term, and r is the common ratio.

We are given that the fifth term T₅ = 96 and the eighth term T₈ = 768. In other words:

$\displaystyle T_5 = a r^{(5) - 1} \text{ and } T_8 = ar^{(8)-1}$

Substitute and simplify:

$\displaystyle 96 = ar^4 \text{ and } 768 = ar^7$

We can rewrite the second equation as:

$\displaystyle 768 = (ar^4) \cdot r^3$

Substitute:

$\displaystyle 768 = (96) r^3$

Hence:

$\displaystyle r = \sqrt[3]{\frac{768}{96}} = \sqrt[3]{8} = 2$

So, the common ratio r is two.

Using the first equation, we can solve for the initial term:

$\displaystyle \begin{aligned} 96 &= ar^4 \\ ar^4 &= 96 \\ a(2)^4 &= 96 \\ 16a &= 96 \\ a &= 6 \end{aligned}$

In conclusion, of the given geometric sequence, the first term a is 6 and its common ratio r is 2.

7 0

2 years ago

I need help with this problem.

DiKsa [7]

Answer:

7 cans will cost $7.29

Step-by-step explanation:

Set up a Proportion: $\frac{24}{25}$ = $\frac{7}{x}$

Cross multiply: 24x=175

Solve for X: x=7.29

5 0

2 years ago