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3 years ago

Please help me!!!!!!!

Mathematics

1 answer:

scZoUnD [109]3 years ago

8 0

Step-by-step explanation:

Write all numerators above the common denominator 7w-5w+6 (w+4)=0 Multiply parenthesis by 6 collect terms 7w-5w+6 (w+4) Wx (w+4) =0 set numerator equal to 0 8w=-24 divide both sides by 8 w=-3

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983488 divided by 1188. What's the remainder? DON'T USE A CALCULATOR!! NO DECIMALS!

matrenka [14]

Answer:

it would be a reminder but it would be C it's the most logical answer because it rounds to a ridiculously large number so it 972 in my opinion

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3 years ago

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2 years ago

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Solve (x – 4)^2=5 Hhhhhbb

Oxana [17]

Answer:

x = 4 + sqrt(5) or x = 4 - sqrt(5)

Step-by-step explanation:

Solve for x:

(x - 4)^2 = 5

Take the square root of both sides:

x - 4 = sqrt(5) or x - 4 = -sqrt(5)

Add 4 to both sides:

x = 4 + sqrt(5) or x - 4 = -sqrt(5)

Add 4 to both sides:

Answer: x = 4 + sqrt(5) or x = 4 - sqrt(5)

6 0

3 years ago

PLEASEE HELPP!!! DUE TODAYY!!!

Alika [10]

Answer:

b 5/12

Step-by-step explanation:

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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

2x-2y=-6

7x =-21

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

2 years ago

Please help me!!!!!!!​

Please help me!!!!!!!