All categories

3 years ago

Adding integers 7th grade

Mathematics

1 answer:

dmitriy555 [2]3 years ago

7 0

Answer:

Integers are a series of positive and negative numbers. When adding such numbers students have to pay attention to the sign that precedes the numbers in the equation. Generally, adding a value to a negative value is like subtracting the negative value from the positive one.

Step-by-step explanation:

You might be interested in

Find the second decile of the following data set 24, 64, 25, 40, 45, 34, 14, 26, 28, 24, 58, 51 D2 =

Sergio039 [100]

Answer:

D2 = 24

Step-by-step explanation:

given is a data set as

24, 64, 25, 40, 45, 34, 14, 26, 28, 24, 58, 51

We have to find the second decile for the above

Arranging in ascending order we get

14, 24, 24, 25,26,28, 34, 40, 45, 51, 58, 64

2nd decile is equal to 20th percentile

20th percentile formula entry appearing which sorts 20% below and 80% above.

i.e. (20%*(14+1)) = 3rd entry

Thus 20th percentile is24.

D2 = 24

7 0

2 years ago

Simplify 4a+3b-(-2a-3b)

Talja [164]

4a+3b-(-2a-3b)
=4a+3b+2a+3b
=6a+6b
=6(a+b)

8 0

3 years ago

Find the slope of a line with points: ( -5 , -8 ) and ( 6 , -3 )

Natasha_Volkova [10]

(y2-y1)/(x2-x1) (-3 - (-8))/(6- (-5))= 5/11

6 0

2 years ago

PLS HURRRRRRRYYYYY TYYY

Ede4ka [16]

Answer:

It's 15.70796. Won't go in-depth since time seems to be of the essence

Step-by-step explanation:

Won't go in-depth since time seems to be of the essence

8 0

2 years ago

Please help me!! 100 points!!

Lyrx [107]

Part A.

We could take, for example:

$\begin{cases}x\geq-3\\y\geq3\end{cases}$

For x ≥ -3 we have vertical line (x-intercept = -3) and shaded region on the right side of the line.
For y ≥ 3 we have horizontal line (y-intercept = 3) and shaded region above that line.

Part B.

Simply, substitute x and y coordinates of D and E <span>to the system of inequalities from part A, and check if its is true, what we get.
</span>
$D=(-2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}-2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}$

so point D is a solution,

$E=(2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}$

and point E is also a solution to <span>our system of inequalities.
</span>
Part C.

There are two ways we could do that. First is the same as in part B. We substitute x and y coordinates of each school and check if inequality is true or false:

$A=(-5,5)\\\\y\ \textless \ 3x-3\quad\implies\quad5\ \textless \ 3\cdot(-5)-3\quad\implies\quad5\ \textless \ -18\quad\text{False}\\\\\\
B=(-4,-2)\\\\y\ \textless \ 3x-3\quad\implies\quad-2\ \textless \ 3\cdot(-4)-3\quad\implies\quad-2\ \textless \ -15\quad\text{False}\\\\\\
C=(2,1)\\\\y\ \textless \ 3x-3\quad\implies\quad1\ \textless \ 3\cdot2-3\quad\implies\quad1\ \textless \ 3\quad\text{True}\\\\\\
D=(-2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot(-2)-3\quad\implies\quad4\ \textless \ -9\quad\text{False}\\\\\\
E=(2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot2-3\quad\implies\quad4\ \textless \ 3\quad\text{False}\\\\\\$

$F=(3,-4)\\\\y\ \textless \ 3x-3\quad\implies\quad-4\ \textless \ 3\cdot3-3\quad\implies\quad-4\ \textless \ 6\quad\text{True}\\\\\\$

So Timothy can attend schools C and F.

We can also draw a graph of that inequality (pic 2).

8 0

2 years ago

Read 2 more answers