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enot [183]
3 years ago
7

For the given expression identify the terms like terms coefficient and constant terms then simplify the expression -8c+-c+1

Mathematics
1 answer:
boyakko [2]3 years ago
6 0

Answer:

The simplified expression is : -9c +1

Step-by-step explanation:

The like terms are -8c and -c

The constant term is 1

The simplified expression is :

-8c + -c + 1 = -9c +1

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A line with a slope of 2/5 passed through the point (-2,3). Write the equations of the line in point-slope form and graph the li
DochEvi [55]

Answer:

y - 3 = (2/5)(x + 2)

Step-by-step explanation:

Start with the general point-slope form of the equation of a straight line:

y - k = m(x - h).  Now substitute -2 for h and 3 for k, and also 2/5 for m:

y - 3 = (2/5)(x + 2)

5 0
3 years ago
Find the zeros of the function. Enter the solutions from least to greatest . h(x) = (- 4x - 3)(x - 3) lesser x = greater
icang [17]

9514 1404 393

Answer:

  -3/4, 3

Step-by-step explanation:

The zeros are the values of x that make h(x)=0. Those are the values of x that make the factors of h(x) be zero.

  -4x -3 = 0   ⇒   x = -3/4

  x -3 = 0   ⇒   x = 3

The zeros of the function are -3/4 and 3.

3 0
3 years ago
What is the correct operation to solve for x
FinnZ [79.3K]
I don’t know I’m sorry
3 0
3 years ago
If the length of a rectangular garden is 2 feet less than the length of a house and the width of the garden is 5 feet less than
Goshia [24]

Answer:

A(h)  =  h²/2 - 6*h + 10

Step-by-step explanation:

The garden is rectangular, and area of a rectangle is:

A(r) = L*w

Where L is the length and w is the width of the rectangle

Now if we call "h" the length of the house, we have the following expressions

L = h - 2  and

w = h/2 - 5

The expression for the area f the ectangle as a function of the length of the house is:

A(h)  =  ( h - 2 )* ( h/2 -5 )

A(h)  = h²/2 - 5*h - h + 10

A(h)  =  h²/2 - 6*h + 10

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3 years ago
Look at the picture attached :)
Karolina [17]
I think it is B
Not sure
4 0
3 years ago
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