All categories

3 years ago

Help me with this someone, please!

Mathematics

2 answers:

Vitek1552 [10]3 years ago

4 0

Answer:

-45 I think

Step-by-step explanation:

6 × (-7) - 3

-42 - 3

-45

Sorry if this is wrong

garri49 [273]3 years ago

4 0

-45 myyyybhomieeeeeee the guy above u is right

You might be interested in

Which of the following points is in the solution set of y < x2 - 2x - 8? (-2, -1)(0, -2)(4, 0)

ratelena [41]

Y < (x - 4)(x + 2)

so the critical points are -2 and 4
(-2,-1) will be in the solution

( 0,-2) will not

(4,0) will not.
the answer is ((-2,-1)

5 0

3 years ago

Read 2 more answers

For every 1/4 mi Pascal walks Lucien runs 2/3 mi. How many miles does Pascal walk for every 1 mi Lucien runs? Write your answers

attashe74 [19]

Answer:

Step-by-step explanation: The answer is 0.5

7 0

3 years ago

Marcy left for school and felt like running. She ran for a while and then stopped to rest. After resting, she walked the rest of

egoroff_w [7]

Answer:

The last graph best represents the story

Step-by-step explanation:

Mary left for school from rest. So the initial speed is zero.

So we start from the origin and draw a straight line with a positive slope to running, that is as time increases , distance covered also increases.

When she is resting, her speed is zero so we draw a horizontal line to indicate this second phase of her journey.

After resting she still continued her journey to school.

This part also has a positive slope.

8 0

3 years ago

Hello can you help me with this problem ?

Lelechka [254]

Step-by-step explanation:

please mark me as brainlist please

3 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