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3 years ago

Solve the following equation for x

Mathematics

2 answers:

OlgaM077 [116]3 years ago

8 0

X-5/2 =-6 multiply both sides
x-5=-12 move the constant to the right
x = -12+5 calcúlate
x=-7

grigory [225]3 years ago

6 0

Answer:

I have no ide- the correct answer -7 pls brainliest

Step-by-step explanation:

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Question 20 Only, Please show your work.<br><br> Solve the problem.

Sveta_85 [38]

Answer:

7 inches

Step-by-step explanation:

The area of a square is given by the formula ...

A = s² . . . . . s = side length

Using the given area, you have ...

49 in² = s² . . . . . fill in given value of area

7 in = s . . . . . . . take the positive square root of both sides

The length of a side of the square is 7 inches.

8 0

3 years ago

The table shows the estimated costs for each semester at the 4-year college Nola plans to attend.

timama [110]

Answer:

Step-by-step explanation:

The estimated cost to attend this college for 1 semester is $9114

3974+400+3240+1500=9114

Nola would have to pay for 8 semesters if she attends for 4 years.

4x2=8

The estimated cost for Nola to attend this college for 4 years is $72912

9114x8=72912

5 0

3 years ago

Read 2 more answers

An urn contains 2n balls, of which 2 are numbered 1, 2 are numbered 2, ... , and 2 are numbered n. balls are successively withdr

madam [21]

She had a 88% chance because she used enough urn

3 0

3 years ago

Express the fifth roots of unity in standard form a + bi. with 1 + 0i

marishachu [46]

Answer:

For K=0

$cos(\frac{0+2\pi 0}{5})+isin(\frac{0+2\pi 0}{5})\\cos(0)+isin(0)=1+i0$

For K=1:

$cos(\frac{0+2\pi 1}{5})+isin(\frac{0+2\pi 1}{5})\\cos(2\pi/5)+isin(2\pi/5)=0.3090+i0.9510$

For K=2:

$cos(\frac{0+2\pi 2}{5} )+isin(\frac{0+2\pi 2}{5} )\\cos(4\pi/5)+isin(4\pi/5)=-0.809+i0.587$

For K=3:

$cos(\frac{0+2\pi 3}{5} )+isin(\frac{0+2\pi 3}{5} )\\cos(6\pi/5)+isin(6\pi/5)=-0.809-i0.587$

For K=4:

$cos(\frac{0+2\pi 4}{5} )+isin(\frac{0+2\pi 4}{5} )\\cos(8\pi/5)+isin(8\pi/5)=0.3091-i0.9510$

Step-by-step explanation:

Fifth Root is given by:

$\sqrt[5]{z}=1+0i$

The above equation will become:

$z=(1+0i)^5$

It can be written as:

$z=[cos(0)+isin(0)]^5$

|z|=1,

According to De-moivre's Theorem:

$z=cos(\frac{0}{5})+isin(\frac{0}{5})\\ z=cos(0)+isin(0)$

Now, Fifth Roots of unity in standard form a + bi :

$\sqrt[5]{z}=[{cos(0+2\pi k)+isin(0+2\pi k)}]^{1/5}$

k=0,1,2,3,4

For K=0

$cos(\frac{0+2\pi 0}{5})+isin(\frac{0+2\pi 0}{5})\\cos(0)+isin(0)=1+i0$

For K=1:

$cos(\frac{0+2\pi 1}{5})+isin(\frac{0+2\pi 1}{5})\\cos(2\pi/5)+isin(2\pi/5)=0.3090+i0.9510$

For K=2:

$cos(\frac{0+2\pi 2}{5} )+isin(\frac{0+2\pi 2}{5} )\\cos(4\pi/5)+isin(4\pi/5)=-0.809+i0.587$

For K=3:

$cos(\frac{0+2\pi 3}{5} )+isin(\frac{0+2\pi 3}{5} )\\cos(6\pi/5)+isin(6\pi/5)=-0.809-i0.587$

For K=4:

$cos(\frac{0+2\pi 4}{5} )+isin(\frac{0+2\pi 4}{5} )\\cos(8\pi/5)+isin(8\pi/5)=0.3091-i0.9510$

6 0

3 years ago

Range of the function f(x) = -x^2 - 5

Naddik [55]

Answer:

Step-by-step explanation:

6 0

3 years ago