Simplify: (−7a − 10c) − (−11a + 20c) − (−24a − 11c)

Select the undefined term that best defines the arrow pictured

rodikova [14]

What is the slope of the line through the point A(-3,4) and B(2,-1)
Best answer
A -1
B 1
C -3
D - 1/3

8 0

3 years ago

What is the mass of an object if it takes a net force of 32 N to accelerate

Sonbull [250]

We will find the answer using the second law of motion i.e. Force is equal to the product of mass and acceleration.

$\sf \: F=ma$

Where,

F is force
M is mass
A is acceleration

In our case,

F = 32 N
A = 0.88m/s
M = ?

Let's solve for M ~

$\tt \: 32 = m \times 0.88$

$\tt \: 32 = 0.88m$

$\tt \: 0.88m = 32$

$\tt \: m = 36.36$

Thus, The mass of object is 36.36 grams...~

5 0

2 years ago

What is the GCF of the numerator and denominator in the following fraction? -99/176

Lady bird [3.3K]

-99 = -9*11
176 = 16*11

Both factorizations have 11 show up. It turns out that 11 is the largest common factor.

Answer: C) 11

Note: if you divide both numerator and denominator by 11, then the fraction reduces to -9/16

8 0

3 years ago

Read 2 more answers

What is the recursive rule for this geometric sequence?

Alekssandra [29.7K]

Answer:

1

Step-by-step explanation:

5 0

2 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

3 years ago