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stellarik [79]
3 years ago
7

How do you evaluate 25 and 3 over 2

Mathematics
1 answer:
Paha777 [63]3 years ago
7 0

Answer:

37.5

Step-by-step explanation:

25 x 3 = 75

75 / 2 = 37.5

25 x 3/2 = 37.5

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Dance club has a $20 start up fee and a $5 monthly fee <br><br> M= <br><br> B=<br><br> Y=
labwork [276]

9514 1404 393

Answer:

  y = 5x +20

Step-by-step explanation:

We assume you want the total cost (y) of x months of membership at the dance club.

This will be the sum of the start-up fee (20) and the product of the monthly fee (5) and the number of months (x).

  y = 5x +20

Comparing this to the slope-intercept form of the equation of a line, we see ...

  y = mx + b

  m = 5

  b = 20

6 0
3 years ago
The new patio at Tonya's house is 5 yards long by 10 feet wide. She is going to cover the patio with square bricks that measure
Klio2033 [76]

Answer:

A

Step-by-step explanation:

Steps to take to determine the answer

  1. Because the length is in yards, it has to be converted to foot
  2. Find the area of the patio using the length and width measured in foot
  3. Determine the area of the square brick
  4. Multiply the cost of the brick by the area

1 yards = 3 foot

5 x 3 = 15 foot

Area of a rectangle = length x breadth

15 x 10 = 150 ft^2

Cost = 150 x 4 = $600

4 0
3 years ago
Y = 4x + 10 y = 3x + 15​
kumpel [21]

Answer:

-(135/31), 60/31

Step-by-step explanation:

y=60/31

x=-(135/31)

5 0
3 years ago
Can someone solve for x
Helen [10]

bearing in mind that, whenever we have an absolute value expression, is in effect a piece-wise function with a positive and a negative version of the expression, so

\bf |x^2-4x-5|=7\implies \begin{cases} +(x^2-4x-5)=7\\\\ -(x^2-4x-5)=7 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ +(x^2-4x-5)=7\implies x^2-4x-5=7\implies x^2-4x-12=0 \\\\\\ (x-6)(x+2)=0\implies x= \begin{cases} 6\\ -2 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ -(x^2-4x-5)=7\implies x^2-4x-5=-7\implies x^2-4x+2=0 \\\\\\ (x-2)(x-2)=0\implies x = 2

6 0
3 years ago
Learning task 2
patriot [66]

The given equations are

2x-3y=-1\cdots(i) \\\\y=x-1\cdots(ii)

(a) Solution by graphing both the equations:

Graph for both the equations (i) and (ii) are in the figure. The point of intersection of both the graph is the solution of the equations.

From the graph, the point of intersection is (4,3)).

Hence, the required solution is (4,3).

(b) Solution by elimination method:

Equations (ii) can be written as -x+y=-1, now multiply it by 2, we have

-2x+2y=-2\cdots(iii)

Add equations (i) and (iii), we have

2x-3y=-1

-2x+2y=-2

__________

\Rightarrow -3y+2y=-1-2\\\\ \Rightarrow -y=-3\\\\ \Rightarrow y=3.

Putting the value of y=3 in equation (i), we have

2x-3\times 3=-1 \\\\\Rightarrow 2x=-1+9=8 \\\\\Rightarrow x=8/2=4

Hence, the required solution is (4,4).

(c) Solution by substitution method:

Substituting the value of y from equation (ii) to equation (i), we have

2x-3(x-1)=-1 \\\\\Rightarrow 2x-3x+3=-1 \\\\\Rightarrow -x=-1-4=-4 \\\\\Rightarrow x=4

Putting the value of x=4 in equation (i), we have

y=4-1=3.

Hence, the required solution is (4,3).

3 0
3 years ago
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