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Veronika [31]
3 years ago
6

Factor completely 3x2 + 9x − 54.

Mathematics
2 answers:
Alinara [238K]3 years ago
7 0
3x2+9x-54

6+9x-54

6-54+9x

-48+9x
MA_775_DIABLO [31]3 years ago
7 0

Answer:

3(x+6)(x-3)

Step-by-step explanation:

3x^2+9x-54

3(x^2+3x-18)

3[(x^2-3x+6x-18)]

3[x(x-3)+6(x-3)]

3[(x+6)(x-3)]

3(x+6)(x-3)

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Which property is used in the equation below?<br> 3(2x + 6) = 6x + 18
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distributive property

Step-by-step explanation:

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3 years ago
Alexander is a stockbroker. He earns 14% commission each week. Last​ week, he sold ​$5,400 worth of stocks. How much did he make
Vlada [557]

Answer:

In that one week, his commision was $896, over all 52 weeks, he made $46592

Step-by-step explanation:

So, he makes 14% commision on any stock he sells. This week, he sold $6,400 worth of stocks, and, hes going to make 14% on it. So, take 6400 and divide that by 14%, and what do you get? $896! So, he made $896 that week. Now, say he made that amount EVERY WEEK. so, all you have to do is know how many weeks are in a year (52) and multiply that by 896, which gives you 46592.

5 0
3 years ago
7. I sell cookies for $2 and muffins for $5. Yesterday, I made $35. I sold a total of 13 items.
ki77a [65]
You sold 10 cookies and 3 muffins.
3 0
3 years ago
Read 2 more answers
Please help, this is 7th grade math
Neporo4naja [7]

Answer:

its -1 and 3/4

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5 0
4 years ago
Read 2 more answers
30. The table below shows the average temperature of a lake in months since April in a certain year. Find
Alchen [17]

Answer:

The temperature, in December is 47 °F

Step-by-step explanation:

The given data for the temperature are;

Month             {}         Temp (°F)

April                   {}          59

May               {}               68

June             {}                71

July                  {}             78

August              {}            74

Let x represent the month, and f(x) = y represent the temperature, we have;

y = a·x² + b·x + c

When x = 1, we have, f(1) = a·1² + b·1 + c = 59

f(1) = a + b + c = 59...(1)

When x = 2, we have, f(2) = a·2² + b·2 + c = 68

f(2) = 4·a + 2·b + c = 68...(2)

When x = 3, we have, f(3) = a·3² + b·3 + c = 71

f(3) = 9·a + 3·b + c = 71...(3)

When x = 4, we have, f(4) = a·4² + b·4 + c = 78

f(3) = 16·a + 4·b + c = 78...(4)

When x = 5, we have, f(5) = a·5² + b·5 + c = 74

f(3) = 25·a + 5·b + c = 74...(5)

Subtracting equation (3) from equation (5), we get;

25·a + 5·b + c - (9·a + 3·b + c) = 74 - 71

16·a + 2·b = 3...(6)

Subtracting equation (1) from equation (2), we get;

4·a + 2·b + c - (a + b + c) = 68 - 59

3·a + b = 9...(7)

Multiply equation (7) by 2 and substract from equation (6) gives;

16·a + 2·b - 2 × (3·a + b) = 3 - 2 × 9

16·a + 2·b - 6·a - 2·b = 3 - 2 × 9

16·a - 6·a + 2·b  - 2·b = 3 - 2 × 9 = -15

10·a = -15

a = -15/10 = -1.5

a = -1.5

From, 3·a + b = 9, we have;

3 × (-1.5) + b = 9

b = 9 + 4.5 =13.5

b = 13.5

From, a + b + c = 59, we have;

-1.5 + 13.5 + c = 59

c = 59 - (-1.5 + 13.5) = 47

c = 47

The quadratic equation becomes, y = a·x² + b·x + c = -1.5·x² + 13.5·x + 47

f(x) = y = -1.5·x² + 13.5·x + 47

For December, we have, x = 9, and f(x) = -1.5×9² + 13.5×9 + 47 = 47

The temperature, in December = 47 °F

8 0
3 years ago
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