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3 years ago

HELP ASPS!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

jasenka [17]3 years ago

7 0

The answer is the first one

irakobra [83]3 years ago

7 0

Answer:

The fourth one

Step-by-step explanation:

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Solve the equation. -3x+2x=8

Andre45 [30]

Answer:

= − 8

Step-by-step explanation:

Combine like terms

Divide both sides of the equation by the same term

Simplify

8 0

3 years ago

4. Two points on a line are (-10, 1) and (5, -5). If

Vsevolod [243]

Answer:

The x-coordinate of another point is zero

Step-by-step explanation:

step 1

Find the slope between the two given points

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

$(-10,1),(5,-5)$

substitute in the formula

$m=\frac{-5-1}{5+10}$

$m=\frac{-6}{15}$

Simplify

$m=-\frac{2}{5}$

step 2

Find the x-coordinate of another point

we have

(x,-3)

we know that

If the other point is on the line, then the slope between the other point and any of the other two points must be the same

Find the slope between the points

$(x,-3),(5,-5)$

Remember that

$m=-\frac{2}{5}$

substitute in the formula

$-\frac{2}{5}=\frac{-5+3}{5-x}$

$-\frac{2}{5}=-\frac{2}{5-x}$

the denominators must be the same

$5=5-x$

$x=0$

therefore

The x-coordinate of another point is zero

3 0

2 years ago

what is the monthly payment for a loan if the amount to finance is 12,385, the APR is 6.9 percent for 5 years and the monthly pa

dimulka [17.4K]

$12,385 APR = 6.9% 5 years= $864.565 monthly payment per $100 is $2.44 $12,385 + $864.565 =$13,222.565/60 =$220.37+ $2.44(2) =$225.25 Monthly

4 0

3 years ago

Read 2 more answers

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

If rstu is a rhombus, then ___________________.

ollegr [7]

a the side lengths must be = but only opposite angles must be = it COULD be a square but is could be many other things.

3 0

3 years ago

Read 2 more answers