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irga5000 [103]
3 years ago
15

Divide Express your answer in scientific notation 10.4 x 10 4 x 10 10.4 x 10 4x10

Mathematics
1 answer:
Fofino [41]3 years ago
8 0
10.4 x 10 = 104
4 x 10 = 40
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What is x-10 when x=11
Romashka-Z-Leto [24]

Answer:

1

Step-by-step explanation:

Because once you know the unknown you can plug it into the equation so now its 11-10 which equals 1

8 0
3 years ago
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If there were 5 million births in a particular country and 3 comma 375 sets of naturally occurring​ triplets, what is the experi
IgorLugansk [536]

Answer:

Probablity =P=6.75*10^-^4

Step-by-step explanation:

The experimental probability can be defined as the ratio of the number of times any particular event has occurred to the total number of times that event has taken place.

Experimental probability = P = No. of event occurrences / total number of occurrences.

In our question statement,

No of naturally occurring triplets are = 3375

Total number of births = 5 million = 5,000,000

Putting values in equation.

P=3375/5000000

P = 0.000675\\P=6.75*10^-^4

4 0
3 years ago
Can someone help me ​
nexus9112 [7]

Answer:

Its C i think

Step-by-step explanation:

3 0
3 years ago
What two products do you get when you cross-multiply the fractions 2/5 and 3/8?
jarptica [38.1K]
Answer: a. 10 and 12

explanation:
4 0
3 years ago
Please help solve this system of equations
stepan [7]

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

Then the system becomes

\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}

Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

\implies u=\pm27

Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

(meaning two solutions are (7, 13) and (-7, -13))

8 0
3 years ago
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