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3 years ago

How to write 2350 million in standard form ?

Mathematics

1 answer:

Arisa [49]3 years ago

6 0

Answer:

200,000,000 + 30,000,000 + 5,00,000

Step-by-step explanation:

you dont need to add the other numbers when you do that and it should be right

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Simplify. (2m3n2)5 10m8n7 32m8n7 32m15n10 10m15n10

Crank

Answer:

10m15n10

Step-by-step explanation:

5x2m=10m

5x3n=15n

5x2=10

so you get 10m15n10

you can't simplify it anymore because none of them have the same variables

5 0

3 years ago

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I know its summer and everyone done with school but someone help

Gemiola [76]

Answer:

1.55

Step-by-step explanation:

3750+900 = 4650

4650/3000 = 1.55

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4 years ago

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Pls help me I’ll give brainliest pls dont answer if you don’t know

Delvig [45]

Answer:

Species

Step-by-step explanation:

i read the passage hence is how i got my answers

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3 years ago

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What does the line 4x -3y = 7 look like?

AlexFokin [52]

C) tilted right upward

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3 years ago

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Which expression represents the number 2i4−5i3+3i2+−81‾‾‾‾√ rewritten in a+bi form?

vichka [17]

Answer:

The expression $-1+14i$ represents the number $2i^4-5i^3+3i^2+\sqrt{-81}$ rewritten in a+bi form.

Step-by-step explanation:

The value of $i$ is $i=\sqrt{-1}[tex] or [tex]i^{2}=-1[\tex].Now [tex]i^{4}$ in term of $i^{2}[\tex] can be written as, [tex]i^{4}=i^{2}\times i^{2}$

Substituting the value,

$i^{4}=\left(-1\right)\times \left(-1\right)$

Product of two negative numbers is always positive.

$\therefore i^{4}=1$

Now $i^{3}$ in term of $i^{2}[\tex] can be written as, [tex]i^{3}=i^{2}\times i$

Substituting the value,

$i^{3}=\left(-1\right)\times i$

Product of one negative and one positive numbers is always negative.

$\therefore i^{3}=-i$

Now $\sqrt{-81}$ can be written as follows,

$\sqrt{-81}=\sqrt{\left(81\right)\times\left(-1\right)}$

Applying radical multiplication rule,

$\sqrt{ab}={\sqrt{a}}\sqrt{b}$

$\sqrt{\left(81\right)\times\left(-1\right)}={\sqrt{81}}\sqrt{-1}$

Now, $\sqrt{\left(81\right)=9$ and $\sqrt{-1}}=i$

$\therefore \sqrt{\left(81\right)\times\left(-1\right)}=9i$

Now substituting the above values in given expression,

$2i^4-5i^3+3i^2+\sqrt{-81}=2\left(1\right)-5\left(-i\right)+3\left(-1\right)+9i$

Simplifying,

$2+5i-3+9i$

Collecting similar terms,

$2-3+5i+9i$

Combining similar terms,

$-1+14i$

The above expression is in the form of a+bi which is the required expression.

Hence, option number 4 is correct.

5 0

3 years ago