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alukav5142 [94]

3 years ago

12

Four times the difference of a number and seven is 32

1 answer:

disa [49]3 years ago

5 0

As a formula, you're saying 4*(n-7)=32

This solves to n-7 = 8, so n=15

You might be interested in

Values of 1/64 check all that apply

xenn [34]

Answer:

there r many values of such fractions just see whether one of it's multiple match with the options like

Step-by-step explanation:

4 0

3 years ago

John read the first 114 pages of a novel, which was 333 pages less than 1/3 of the novel. Write an equation to determine the tot

tankabanditka [31]

(p)=(333+114)•3

because 114 pages are already read and if he needs 333 pages to reach 1/3 of the book you add 114 and 333 so you can find out how much pages is 1/3 of the book. then you multiply that number by 3 because there are 3 sections of 333+114.

3 0

3 years ago

Suzannes school are painting a rectangular mural outside the buillding that will be 15 feet by 45 feet the students are also goi

tatuchka [14]

Answer:

THE LENGTH OF THE STUDENTS PAINTING=15/3=5

THE BREADTH OF THE STUDENTS PAINTING=45/3=15

AREA OF THEIR PAINTING=15×5=75

4 0

3 years ago

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

(8x - 14) (5y + 16) (5x + 34)

labwork [276]

Answer: 200x^2y+640x^2+1010xy+3232x-2380y-7616

Step-by-step explanation:

((8x−14)(5y+16))(5x+34)

((8x−14)(5y+16))(5x)+((8x−14)(5y+16))(34)

200x2y+640x2−350xy−1120x+1360xy+4352x−2380y−7616

200x2y+640x2+1010xy+3232x−2380y−7616

3 0

3 years ago