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velikii [3]
3 years ago
13

The quotient of a number and negative three, increased by two, is thirty nine

Mathematics
1 answer:
Leokris [45]3 years ago
4 0
(x/-3)+2=39
minus 2 both sides
(x/-3)=37
times both sides by -3
x=-111
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Zapisz w jak najprostszej postaci:<br> a) -x-4(1-x)<br> b) 3(5x-2)-5(3-4x)<br> c) √2(3-x)-2(1-x√2)
Sonja [21]

Answer:

Step-by-step explanation:

a) -x-4(1-x)  =

− 2 x −4 ( 1 − x ) .

2 x − 4

b) 3(5x-2)-5(3-4x)  =

3 ⋅ (5 x − 2 ) − 5 ⋅ ( 5 − 4 x )

35 x  −  31

c) √2(3-x)-2(1-x√2) =

√ 2 x ⋅ ( 3 −x ) − 2 ⋅( 1 − x √ 2 ) .

2 x √ 2 + 3 √ 2 x -√ 2 x x − 2

4 0
3 years ago
Help! Best answer gets brainliest
julsineya [31]

Answer:

7.9

Step-by-step explanation:

the first thing to do is find the second projection.

16-7 = 9

than we can use a proportion to find BD

7 : BD = BD : 9

BD = √9 x 7 = √63 = 7.9

we have to use the square root because in the proportion BD Is repeated two times. If we don’t use it, we find BD^2

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=z%5E%7B7%7D%3D128i" id="TexFormula1" title="z^{7}=128i" alt="z^{7}=128i" align="absmiddle" cla
abruzzese [7]

If <em>z</em> ⁷ = 128<em>i</em>, then there are 7 complex numbers <em>z</em> that satisfy this equation.

z^7 = 128i = 2^7i = 2^7e^{i\frac\pi2}

\implies z=\sqrt[7]{2^7} e^{i\frac17\left(\frac\pi2+2n\pi\right)}

(where <em>n</em> = 0, 1, 2, …, 6)

\implies z = 2 e^{i\left(\frac\pi{14}+\frac{2n\pi}7\right)}

\displaystyle\implies z = 2 \left(\cos\left(\frac\pi{14}+\frac{2n\pi}7\right)+i\sin\left(\frac\pi{14}+\frac{2n\pi}7\right)\right)

3 0
3 years ago
Solve for n: 3 x n = 2 x 9<br> I need help
Nikolay [14]

Answer:

18

Step-by-step explanation:

CALCULATER

4 0
3 years ago
Read 2 more answers
Evaluate the expression mentally,<br> 35 * 91 - 35 * 89
-Dominant- [34]

Answer:

2870

Step-by-step explanation: mind

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2 years ago
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