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Nikitich [7]
3 years ago
8

Join the zoommmmmmmmmmmmml

Mathematics
2 answers:
eduard3 years ago
8 0

do u need help on joining a zoom i can help i zoon everyday

amm18123 years ago
5 0

Answer:

The question is imcomplete.

Step-by-step explanation:

There is no zoom link or real question.

You might be interested in
Pls help if u can 6th grademath
omeli [17]

Answer:

n = 54

Step-by-step explanation:

\frac{2}{12}=\frac{9}{n}\\

Cross products,

2 * n = 9 *12

 2n = 108

Divide both sides by 2

n = \frac{108}{2}\\

n = 54

4 0
3 years ago
Expand the given power using the Binomial Theorem. (10k – m)5
agasfer [191]

Answer:

(10k - m)^{5}=100000k-50000k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}

Step-by-step explanation:

* Lets explain how to solve the problem

- The rule of expand the binomial is:

(a+b)^{n}=(a)^{n}+nC1(a)^{n-1}(b)+nC2(a)^{n-2}(b)^{2}+nC3(a)^{n-3}(b)^{3}+...............+(b)^{5}

∵ The binomial is (10k-m)^{5}

∴ a = 10k , b = -m and n = 5

∴ (10k-m)^{5}=(10k)^{5}+5C1(10k)^{4}(-m)+5C2(10k)^{3}(-m)^{2}+5C3(10k)^{2}(-m)^{3}+5C4(10k)^{1} (-m)^{4}+5C5(10k)^{0}(-m)^{5}

∵ 5C1 = 5

∵ 5C2 = 10

∵ 5C3 = 10

∵ 5C4 = 5

∵ 5C5 = 1

∴ (10k-m)^{5}=100000k^{5}+(5)(10000)k^{4}(-m)+(10)(1000)k^{3}(m^{2})+(10)(100)k^{2}(-m^{3})+5(10k)^{1} (m^{4})+(10k)^{0}(-m^{5})

∴ (10k-m)^{5}=100000k^{5}-50000)k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}

5 0
3 years ago
The radioactive substance cesium-137 has a half-life of 30 years. The amount At (in grams) of a sample of cesium-137 remaining a
nadya68 [22]

Answer:

Initial amount is 523 grams

Amount after 100 years is 52 grams

Step-by-step explanation:

General exponential function for radioactive substances is given by

A(t) = Ao(1/2)^t/t1/2

Where,

A(t) is the amount remaining after t years

Ao is the initial amount of the radioactive substance

t is the number of years required for the radioactive substance to decay to A

t1/2 is the half-life of the radioactive substance

Comparing A(t) = 523(1/2)^t/30 with the general exponential function

Initial amount (Ao) = 523 grams

A(t) = 523(1/2)^t/30

t = 100

A(100) = 523(1/2)^100/30 = 523×0.5^3.333 = 523×0.0992 = 51.8816 = 52 grams (to the nearest gram)

7 0
3 years ago
Count and fill in the missing numbers.<br> ..., 44, 45, 46, 47, 48, <br> , <br> , <br> , ...
Alecsey [184]

Answer:

43

Step-by-step explanation:

That is the answer

4 0
3 years ago
Read 2 more answers
Properties of Logarithms pls help me with this?
miskamm [114]

Answer:

log 18= 2.0850

1.2925+0.7925= 2.085

5 0
3 years ago
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