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arlik [135]
2 years ago
11

I really need help with this, please help me.

Mathematics
1 answer:
vazorg [7]2 years ago
3 0

Answer:

See below.

Step-by-step explanation:

1. h

2. g

3. a

4. f

5. c

6. b

7. e

8. (17 * 5) * 2 = 17 * (5 * 2) = 17 * 10 = 170

9. 700 + 137 + 300 = 700 + 300 + 137 = 1137

10. $0.25 + $2.69 + $4.75 = $0.25 + $4.75 + $2.69 = $7.69

11. -8 + 57 + 18 = 18 - 8 + 57 = 10 + 57 = 67

12. 26 + 19 + 14 = 20 + 6 + 19 + 10 + 4 =

= 20 + 10 + 6 + 4 + 19

= 30 + 10 + 10

= 40 + 19

= 59

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6x minus 2y equals 24
Drupady [299]

Answer:

slope:-3 y-intercept 12

6 0
3 years ago
Which expression represents the product of 3 and y? 3 + y 3 − y 3 over y 3y
Eva8 [605]

Answer:

3y is the expression that represents the product of 3 and y

5 0
3 years ago
How do I make fractions into mixed numbers??? like 3 8/4 ​
prohojiy [21]

Answer:

look down

Step-by-step explanation:

OK so u take something like 9/3 and then ask how many times the 3 goes into the 9 and that is your whole number. whatever is left is your top of the fraction and the bottom stays the same hope this helped

8 0
2 years ago
Read 2 more answers
What are the roots of this equation? x2-4x+9=0
zysi [14]

Considering the definition of zeros of a function, the zeros of the quadratic function f(x) = x² + 4x +9 do not exist.

<h3>Zeros of a function</h3>

The points where a polynomial function crosses the axis of the independent term (x) represent the so-called zeros of the function.

In summary, the roots or zeros of the quadratic function are those values ​​of x for which the expression is equal to 0. Graphically, the roots correspond to the abscissa of the points where the parabola intersects the x-axis.

In a quadratic function that has the form:

f(x)= ax² + bx + c

the zeros or roots are calculated by:

x1,x2=\frac{-b+-\sqrt{b^{2}-4ac } }{2a}

<h3>This case</h3>

The quadratic function is f(x) = x² + 4x +9

Being:

  • a= 1
  • b= 4
  • c= 9

the zeros or roots are calculated as:

x1=\frac{-4+\sqrt{4^{2}-4x1x9 } }{2x1}

x1=\frac{-4+\sqrt{16-36 } }{2x1}

x1=\frac{-4+\sqrt{-20 } }{2x1}

and

x2=\frac{-4-\sqrt{4^{2}-4x1x9 } }{2x1}

x2=\frac{-4-\sqrt{16-36 } }{2x1}

x2=\frac{-4-\sqrt{-20} }{2x1}

If the content of the root is negative, the root will have no solution within the set of real numbers. Then \sqrt{-20} has no solution.

Finally, the zeros of the quadratic function f(x) = x² + 4x +9 do not exist.

Learn more about the zeros of a quadratic function:

brainly.com/question/842305

brainly.com/question/14477557

#SPJ1

6 0
1 year ago
The nth term of a sequence is n + 5<br> Work out the 5th term of the sequence.<br> 5th term =
Ilya [14]

Answer:

10

Step-by-step explanation:

Tₙ=n+5

T₅=5+5=10

4 0
3 years ago
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