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qwelly [4]
3 years ago
5

Find the missing coordinate to complete the ordered pair.

Mathematics
2 answers:
Illusion [34]3 years ago
6 0

Answer:

<u>(</u><u>9</u><u>,</u><u> </u><u>-</u><u>4</u><u>)</u>

Step-by-step explanation:

{ \bf{y =  - x + 5 \:  \: (9, - )}}

Substitute x with 9, to find x:

{ \sf{y =  - 9 + 5}} \\ y =  - 4

nordsb [41]3 years ago
3 0

Answer:

(9, -4)

Step-by-step explanation:

Substitute 9 for  x  in the expression.

y = -(9) + 5

y = -9 + 5

y = -4

The ordered pair is (9, -4).

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Which point gives the vertex of ƒ(x) = –x2 + 4x – 3?
Dmitriy789 [7]

Answer:

The vertex is (2,1)

Step-by-step explanation:

ƒ(x) = –x^2 + 4x – 3

Factor out the negative

   = -(x^2 -4x+3)

 Factor

What 2 numbers multiply to +3 and add to -4

-3*-1 = 3

-3+-1 = -4

f(x) =  -( x-3)(x-1)

Find the zeros

0 = -( x-3)(x-1)

0 = x-3   0 = x-1

x=3            x=1

The x value of the vertex is 1/2 way between the two zeros

(3+1)/2 = 4/2 =2

To find the y value, substitute x=2 in

f(2) =  -( 2-3)(2-1)

       =-(-1)(1) = 1

The vertex is (2,1)

4 0
3 years ago
Nick's Printing Press promptly sold 5-cent pencils and 10-cent pens to the public. It sold 1000 writing utensils and took $74.50
galben [10]

Answer:

  • 490 pens
  • 510 pencils

Step-by-step explanation:

Let p represent the number of pens Nick sold. Then the revenue is ...

  0.10p +0.05(1000 -p) = 74.50

  0.05p = 24.50 . . . . . . eliminate parentheses, subtract 50

  p = 490 . . . . . . . . . . . . multiply by 20

The number of pencils sold is then ...

  pencils = 1000 -490 = 510

Nick's Printing Press sold 490 pens and 510 pencils.

4 0
3 years ago
Question 2 please fas
Len [333]

Answer:

a = 3, b = 2

Step-by-step explanation:

It is given that both the polynomials are equal.

Therefore, h(x) = k(x)

{x}^{3}  + (a + b) {x}^{2}  - 4x + 2 \\  =  {x}^{3}  + 5 {x}^{2}  - (2a - b)x + 2 \\  \\ equating \: like \: terms \: on \: both \: sides \\ (a + b) {x}^{2}  = 5 {x}^{2}   \\  \implies \: a + b = 5.....(1) \\  \\  - (2a - b)x =  - 4x \\ 2a - b = 4....(2) \\  \\ adding \: equations \: (1) \: and \: (2) \\  \\ 3a = 9 \\   \implies \: a =  \frac{9}{3}  \\   \huge \red{ \boxed{\implies \: a = 3}} \\  \\ substituting \: a = 3 \: in \: equation \: (1) \\ 3 + b = 5 \\   \implies \: b = 5 - 3 \\ \huge \purple{ \boxed{ \implies  b = 2}}

5 0
3 years ago
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