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jekas [21]
4 years ago
12

35 POINTS. I need help! Please!

Mathematics
1 answer:
topjm [15]4 years ago
7 0
There is no solution  for the first one

if you eliminate y   you get 2 equations
-13x - 13z = -25
-13x - 13x = -15    

- there is no solution to theses

a23  means the element in the second  row and the 3rd column

so its -5 
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The difference between 7 7/8 - 3 1/4
Dima020 [189]
You could convert them into improper fractions first. 7 7/8 should become 63/8 and 3 1/4 should become 13/4. Then, change both fractions so they share a common denominator. I'll use 8. 
63/8-26/8 = 37/8, which is then converted into 4 5/8.
Hope that helped you.
8 0
3 years ago
Read 2 more answers
Lynn plotted point G, 3 units to the left and 2 units above point F . Where did Lynn plot point G ?
bekas [8.4K]

Answer:

(2, 6)

Step-by-step explanation:

Point G has a coordinate of x = 5, and y = 4, that is (5, 4).

If Lynn plots point G, such that:

G is 3 units to the left of point F, the x-coordinate of point G = 5 - 3 = 2

G is 2 units above point F, the y-coordinate of point G = 4 + 2 = 6.

Therefore, Lynn plotted point G at x = 2, and y = 6. Which is (2, 6)

3 0
3 years ago
At a local fitness​ center, members pay a ​$10 membership fee and ​$5 for each aerobics class. Nonmembers pay ​$6 for each aerob
sveta [45]
One way to solve it is writing two equations based on the info so

y = 5x + 10 for members
y = 6x for nonmembers

(the + 10 isn't included in the second equation because nonmembers won't pay a membership fee)

and to find their intersection point (where the values will be the same), you just set them equal to each other:

5x + 10 = 6x

subtract 5x

x = 10 classes
7 0
3 years ago
Solve for x. enter your answer in interval notation using grouping symbols. x^2+15x<-36
Oliga [24]
X²+15x+36<0

at first solve quadratic equation

D=b²-4ac= 225-4*1*36= 81

x=(-b+/-√D)/2a
x=(-15+/-√81)/2= (-15+/-9)/2
x1=(-15-9)/2=-12
x2=(-15+9)/2=-3

we can write x²+15x+36<0 as (x+12)(x+3)<0

(x+12)(x+3)<0 can be 2 cases, because for product to be negative one factor should be negative , and second factor should be positive
 1 case)      x+12<0, and x+3>0,                                            
                       x<-12, and x>-3
(-∞, -12) and(-3,∞) gives empty set

or second case)  x+12>0 and x+3<0
x>-12 and x<-3
(-12,∞) and (-∞,-3)  they are crossing , so (-12, -3)  is a solution of this inequality


3 0
4 years ago
How do you graph a possible situation?
Svet_ta [14]
To graph a situation that would involve a linear graph, first determine your x and y axes.
The x-axis will be the independent variable, one that does not change based on other variables. An example is time.

The y-axis, the dependent variable, depends on the independent variable.

The model equation for a linear line is y = mx + b.
"m" is the slope, and the "b" is the y-intercept (where the graph crosses the x-axis at x=0).

For example, a situtation could be that Joe starts with $10 in his account and adds $5 every day to his account.
The x-axis is time in days.
The y-axis is amount of money in his account.
The slope, or rate of change is 5.
The y-intercept, the amount of money he has at x=0 (0 days) is $10.

The equation would be y = 5x + 10
To draw this, plot the y-intercept at (0, 10), and the next point would be 5 units up and one unit to the right because the slope is 5, or 5/1 (remember slope is rise over run: "rise" up 5 and "over" to the right 1).

4 0
3 years ago
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