Can someone help me asap.

Mathematics

2 answers:

s2008m [1.1K]4 years ago

6 0

Answer:

option 1

Step-by-step explanation:

$\sqrt{y} =y^{\frac{1}{2}}\\\\(\frac{1}{\sqrt{y} })^{\frac{-1}{5}} =(\frac{1}{y^{\frac{1}{2}}})^{\frac{-1}{5}}\\\\= (y^{\frac{-1}{2}})^{\frac{-1}{5}}\\\\=y^{\frac{-1}{2}*\frac{-1}{5}}\\\\= y^{\frac{1}{10}}\\\\=\sqrt[10]{y}$

dalvyx [7]4 years ago

4 0

Answer:

$- \frac{1}{5 \sqrt{y} }$

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jekas [21]

Answer:

y= -12
x = 7
q = -1
x = 24

Step-by-step explanation:

1. ..............................................

6(y + 7) = 2(y - 3)
6y + 42 = 2y - 6
6y - 2y = -6 - 42
4y = -48
y = -12

2...............................................

3x + 2(x - 5) = 25
3x + 2x - 10 = 25
5x = 25 + 10
5x = 35
x = 7

3 . ..............................................

10q + 3q + 5 = 2(q - 3)
13q + 5 = 2q - 6
13q - 2q = -6 - 5
11q = -11
q = -1

4. ..............................................

1/3(x + 6) = 2/3(x - 9)
3*1/3(x+6) = 3*2/3(x - 9)
x + 6 = 2(x - 9)
x + 6 = 2x - 18
2x - x = 6 + 18
x = 24

3 0

3 years ago

Read 2 more answers

What is 33 3/1 as a decimal and how do you change it to decimal?

Leto [7]

Decimal: 37

You leave 33 as it is since it is a whole number. Then you divide 3 ÷ 1 because the fraction is $\frac{3}{1}$ . 3 ÷ 1 = 3
33+3 = 37

#7 is your answer.

6 0

4 years ago

Find a third-degree polynomial equation with the rational coefficients that has roots -3 and 1+i

jolli1 [7]

Answer:

x³ + x² − 4x + 6 = 0

Step-by-step explanation:

Imaginary roots come in conjugate pairs. So if 1+i is a root, then 1−i is also a root.

(x − (-3)) (x − (1+i) (x − (1−i)) = 0

(x + 3) (x² − (1+i) x − (1−i) x + (1+i) (1−i)) = 0

(x + 3) (x² − x − ix − x + ix + 1 − i²) = 0

(x + 3) (x² − 2x + 2) = 0