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1 year ago

A bank charges a service fee of 70.50 per month for a checking acount

Mathematics

1 answer:

timurjin [86]1 year ago

4 0

The months till the account stability is terrible is 1.3 months.

<h3>What is account stability?</h3>

In banking, the account stability is the amount of coins you have were given available on your checking or economic financial savings account.
Your account stability is the net amount available to you anyways deposits and credit score had been balanced with any costs or debits.
Your economic organization account balance suggests you methods a brilliant deal coins you have were given on your account.
That coins is there for you withdraw or depart in place, likely with a view to collect interest payments on it.
The answer isn't always truely yes, though, because of the truth your balance might not be exactly what it seems.
The available balance can be taken out of the account in cash at an ATM or with a economic organization teller.
The debit card transfers coins from the coins inside facet the checking account.

To learn more about account stability from the given link:

brainly.com/question/13387881

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the temperature was 15 .it dropped so that the temperature was 0 . what integer represents the change in temperature?

Nataliya [291]

Answer:

if the temperature was at 15 and dropped to zero the amount it changed would be is -15. The integer is -15 because an integer is a whole number and it can also be a negative or positive number it just can not be a fraction

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3 years ago

Help me please !!!!!!!

KATRIN_1 [288]

Answer:

m<ACB = 107*

Step-by-step explanation:

First we have to find the angle of m<CEB.

We know that the value of that angle plus m<CEP equals a total of 180, so we subtract the value that we have (147*) from 180 and we get 33*.

33* Is also the value of m<CAB.

33+40 = 43. 180-43 = 107.

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2 years ago

The number of hours you study for an exam affects your grade. What is a reasonable value of the range?

Gwar [14]

Answer:

Step-by-step explanation:

The range is the grade. Domain is the amount of hours that you study for.

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3 years ago

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QOTD: What's 658x923? Do NOT use a calculator. Whoever, answers this correctly, I will give BRAINLIEST.

Helga [31]

Answer:

the awnser is 607,334 because when you multiply by 3 its the same like multipliying on

Step-by-step explanation:

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2 years ago

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Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

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