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3 years ago

How do i delete a question

Mathematics

1 answer:

borishaifa [10]3 years ago

4 0

Answer:

I'm almost sure you can't, I also tried but I see no dots or options.

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Please help me now I’m a beginner and I so need help please explain how did you get you answer so I understand thanks

Irina-Kira [14]

Answer:y=2.5(x)

Step-by-step explanation:y=2.5(x)

y=2.5(2)= y=5

7 0

3 years ago

A number added to itself is one

Ivenika [448]

The number would be .5 I'm pretty sure

6 0

3 years ago

Help me out smartie pants hahahahaahahaha

MAXImum [283]

Answer:

5/8

I had it

4 0

3 years ago

1)A System of equations is shown below . What is the solution to the system of equations? 5x+2y=-15 2x-2y=-6

meriva

Answer:

x= -3 and y= 0

Step-by-step explanation:

5x+2y=-15

x= -3

Putting value of x in equation 1

5(-3) +2y=-15

-15+2y= -15

2y= 0

y= 0

This can be solved with the help of matrices

In matrix form the above equations can be written in the form

$\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$

Let

$\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]$ = A $\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = X and $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$ = B

Then AX= B

or X= A⁻¹ B

where A⁻¹= adj A/ ║A║ where mod A≠ 0

adj A= $\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]$

║A║= ( 5*-2- 2*2)= -10-4= -14≠0

X= A⁻¹ B

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]$ $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}42\\0\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-3\\0\\\end{array}\right]$

From here x= -3 and y= 0

Solution Set = [(-3,0)]

3 0

3 years ago

Hi can anyone help me with this ..

aleksley [76]

Answer:

n = π/6, π/4, 3π/4, 5π/6

Step-by-step explanation:

sin(3n) − sin n = cos(2n)

Use double and triple angle formulas:

(3 sin n − 4 sin³ n) − sin n = 1 − 2 sin² n

2 sin n − 4 sin³ n = 1 − 2 sin² n

4 sin³ n − 2 sin² n − 2 sin n + 1 = 0

Factor by grouping:

2 sin² n (2 sin n − 1) − (2 sin n − 1) = 0

(2 sin² n − 1) (2 sin n − 1) = 0

Solve:

2 sin² n − 1 = 0

2 sin² n = 1

sin² n = 1/2

sin n = √2/2

n = π/4, 3π/4

2 sin n − 1 = 0

2 sin n = 1

sin n = 1/2

n = π/6, 5π/6

4 0

3 years ago