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3 years ago

What is 30 x 45 x 34 x89

Mathematics

1 answer:

Finger [1]3 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

Give an example how to convert <br> $8.50 multiplied by 8 equal 68

tatuchka [14]

If you're looking for a word problem here's one,

Nichole has $8.50 and she decides to invest it in to a company worth $1,000,000 in a year the company is worth $8,000,000 as the company went up in value so did Nichole's investment, how much was Nichole's Investment worth at the end of that year?

If your'e looking for someone to show the math here's that,

if you take the decimal first and multiply that by 8 you get 4 because 0.5*8=4, then you multiply the whole number and get 64 because 8*8=64, add the two together and you get 68, 64+4=68

6 0

4 years ago

Is it possible to have a playground area made from 70 yards of fencing with an area of 324 square yards? Explain with

cricket20 [7]

Answer:

Yes: The answer would be 113.04 cubic feet

7 0

3 years ago

A graphic charges a fee of 1600 to design a poster for a jazz festival the designer also receives 8% commission on sales of the

iris [78.8K]

Answer:

The total income of designer from this work is $1649.6.

Step-by-step explanation:

The fees charged by the designer = $1600

The commission on total sales = 8%

Total sales amount in the festival = $620

Now, 8% of $620 = $\frac{8}{100} \times 620 = 49.6$

So, the commission on total sales = $49.6

Also, Total Income of the designer = Deign Fee + 8% Commission fee

= $1600 + $49.6

= $1649.6

Hence, the total income of designer from this work is $1649.6.

5 0

4 years ago

Radioactive Decay

SpyIntel [72]

Answer:

Percentage of (226Ra) after 900 years is 68%

Step-by-step explanation:

Let P(t) be the amount of (226Ra) present at any time t

Half life of (226 Ra) = 1599 years

If P₀ is initial amount of (226 Ra) then after 1599 years

P(1599)=P₀/2

Decay i amount of radioactive substance is related to time t as

$\frac{dP}{dt}=kP(t)\\\\\frac{1}{P}\,dP=kdt\\\\Integrating\,\, both\,\,sides\\\\ln|P|=kt+c\\\\P(t)=Ce^{kt}\\\\at \,\, t=0\,\, P(0)=P_{o}\\\\P(0)=Ce^{k0}\\\\P_{o}=C\\\\then\\\\P(t)=P_{o}e^{kt}$

To find value of k

$at\,\, t=1599\,years\\\\P(1599)=\frac{P_{o}}{2}\\\\then\\\\\frac{P_{o}}{2} =P_{o}e^{k(1599)}\\\\\frac{1}{2} =e^{k(1599)}\\\\ln|\frac{1}{2}|=k(1599)\\\\k=\frac{ln|\frac{1}{2}|}{1599}=-4.3\times 10^{-3}\\\\\implies P(t)=P_{o}e^{-4.3\times 10^{-3}t}\\\\at\,\, t=900 \\\\P(900)=P_{o}e^{-4.3\times 10^{-3}(900)}\\\\P(900)=0.68P_{o}$

Percentage of radioactive element is:

Amount after 900 years $=\frac{P(900)}{P_{o}}\times 100\\\\=\frac{0.68P_{o}}{P_{o}}\times 100\%\\\\=68\%$

3 0

3 years ago

(2^8 ⋅ 5^−5 ⋅ 19^0)^−2 ⋅ 5 to the power of negative 2 over 2 to the power of 3, whole to the power of 4 ⋅ 2^28 (5 points)

DENIUS [597]

Answer:

the answer is 400

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers