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3 years ago

Perform the indicated operation and write the answer in the form a + bi. (–1 + 5i) + (3 + 7i)

Mathematics

1 answer:

pashok25 [27]3 years ago

5 0

Answer:

12i + 2

Step-by-step explanation:

Combine like terms.

-1 + 3 = 2

5i + 7i = 12i

12i + 2 is your answer.

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A pack of clothes pins has 1 quarter yellow 2 thirds red and the rest blue how much blue pins are there

IgorC [24]

Answer: 1/12

Step-by-step explanation:

First make all the fractions have the same denominator. To do that, multiply the fractions by a number that will get them the same denominator. In this case, multiply 1/4 by 3, and 2/3 by 4.

1/4 x 3/3 = 3/12

2/3 x 4/4 = 8/12

then you have 3/12 + 8/12 = 11/12

so, blue is 1/12

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2 years ago

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A phone company offers two monthly plans. Plan A costs 14 plus an additional 0.09 for each minute of calls. Plan B costs 18 plus

Sliva [168]

Answer:

200 minutes

Step-by-step explanation:

14 + 0.09 x = 18 + 0.07 x

0.09 x - 0.07 x = 18 - 14

0.02 x = 4

x = 400/2 = 200

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2 years ago

Use trigonometric substitution to evaluate the integral 13 + 12x − x2 dx . First, write the expression under the radical in an a

SCORPION-xisa [38]

I don't see a square root sign anywhere, so I'll assume the integral is

$\displaystyle\int\sqrt{13+12x-x^2}\,\mathrm dx$

First complete the square:

$13+12x-x^2=49-(6-x)^2=7^2-(6-x)^2$

Now in the integral, substitute

$6-x=7\sin t\implies\mathrm dx=-7\cos t\,\mathrm dt$

so that

$t=\sin^{-1}\left(\dfrac{6-x}7\right)$

Under this change of variables, we have

$7^2-(6-x)^2=7^2-7^2\sin^2t=7^2(1-\sin^2t)=7^2\cos^2t$

so that

$\displaystyle\int\sqrt{13+12x-x^2}\,\mathrm dx=-7\int\sqrt{7^2\cos^2t}\,\cos t\,\mathrm dt=-49\int|\cos t|\cos t\,\mathrm dt$

Under the right conditions, namely that cos(t) > 0, we can further reduce the integrand to

$|\cos t|\cos t=\cos^2t=\dfrac{1+\cos(2t)}2$

$\displaystyle-49\int|\cos t|\cos t\,\mathrm dt=-\frac{49}2\int(1+\cos(2t))\,\mathrm dt=-\frac{49}2\left(t+\frac12\sin(2t)\right)+C$

Expand the sine term as

$\dfrac12\sin(2t)}=\sin t\cos t$

Then

$t=\sin^{-1}\left(\dfrac{6-x}7\right)\implies \sin t=\dfrac{6-x}7$

$t=\sin^{-1}\left(\dfrac{6-x}7\right)\implies \cos t=\sqrt{7^2-(6-x)^2}=\sqrt{13+12x-x^2}$

So the integral is

$\displaystyle-\frac{49}2\left(\sin^{-1}\left(\dfrac{6-x}7\right)+\dfrac{6-x}7\sqrt{13+12x-x^2}\right)+C$

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2 years ago

12+6 What is the value of when N= 2? N O A. 6 OB. 9 O C. 12 O D. 18

Artemon [7]

Answer:

Where is the N in the equation

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3 years ago

PLA HELP QUESTION IN PHOTO

asambeis [7]

Answer:

Step-by-step explanation:

its the first one

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2 years ago

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