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3 years ago

What is the value of xa. 25b. 12c. 7 d. 5

Mathematics

2 answers:

tia_tia [17]3 years ago

8 0

The answer is 5 because a^2 + b^2 = c^2

Ipatiy [6.2K]3 years ago

4 0

4 times 3 is 12 so the answer will be 12 I think

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It wont be a perfect number

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3 years ago

HELP?!??!??!!??!!!!?

larisa86 [58]

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Step-by-step explanation:

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3 years ago

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An engineer deposits $2,000 per month for four years at a rate of 24% per year, compounded semi-annually. How much will he be ab

Alex17521 [72]

The amount of money he will be able to withdraw after 10 years after his last deposit is $926,400.

<h3>Compound interest</h3>

Principal, P = $2,000 × 12 × 4

= $96,000

Time, t = 10 years
Interest rate, r = 24% = 0.24
Number of periods, n = 2

A = P(1 + r/n)^nt

= $96,000( 1 + 0.24/2)^(2×10)

= 96,000 (1 + 0.12)^20

= 96,000(1.12)^20

= 96,000(9.65)

= $926,400

Therefore, the amount of money he will be able to withdraw after 10 years after his last deposit is $926,400

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brainly.com/question/24924853

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2 years ago

What is the max value of x and the min value of x???

LUCKY_DIMON [66]

Max value of x is infinite and min value of x is 0

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3 years ago

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Find e^cos(2+3i) as a complex number expressed in Cartesian form.

ozzi

Answer:

The complex number $e^{\cos(2+31)} = \exp(\cos(2+3i))$ has Cartesian form

$\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2)$ .

Step-by-step explanation:

First, we need to recall the definition of $\cos z$ when $z$ is a complex number:

$\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}$ .

Then,

$\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}$ . (I)

Now, recall the definition of the complex exponential:

$e^{z}=e^{x+iy} = e^x(\cos y +i\sin y)$ .

So,

$e^{2i-3} = e^{-3}(\cos 2+i\sin 2)$

$e^{-2i+3} = e^{3}(\cos 2-i\sin 2)$ (we use that $\sin(-y)=-\sin y)$ .

Thus,

$e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)$

Now we group conveniently in the above expression:

$e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2$ .

Now, substituting this equality in (I) we get

$\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2$ .

Thus,

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)$

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right]$ .

5 0

3 years ago

What is the value of xa. 25b. 12c. 7 d. 5​

What is the value of xa. 25b. 12c. 7 d. 5