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nata0808 [166]
4 years ago
11

Please please please help me with this answer

Mathematics
1 answer:
kow [346]4 years ago
3 0

There are 3 girls .

9 students , 3 boys & 3 girls + another 3 boys because there are 3 more than girls so there are only 3 girls since you can't exceed 9 .

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PLEASE I NEED HELP NOW!!!!
const2013 [10]

Answer:

80

Step-by-step explanation:

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5 0
3 years ago
Read 2 more answers
8 to power -2 multiplied by a to power of 7
levacccp [35]
<span> (8^ -2)(4^4)(16^ -1)

</span>
<span>a^-x=(1/a^x)

</span>
<span> (4^4)/((8^2)(16))

"NOTE"  </span><span>16^1=16

</span>
<span>4^4=256

 8^2=64

256/((64)(16))=1/4
</span>
<span>convert 1/4 to a power of two
</span>
<span>(4) is 2

 1/4=1/(2^2)
</span>
<span>1^2=1

 (1/2)^2=(1/4)

</span>
<span>1/2 to the power of 2 is the answer

It would look like this

1/2²
</span>
8 0
3 years ago
Helpppppppppppppppppppppp
Ivenika [448]
Answer:
-1

Explanation:
3π / 4 = (3*180) / 4 = 135 degrees
Since 135 is in the 2nd quadrant, therefore:
tan (135) = - tan (45) = -1

cot θ = 1/tanθ
Therefore:
cot (3π / 4) = 1/tan(3π / 4)
cot(3π / 4) = -1

Hope this helps :)
5 0
3 years ago
The temperature dropped at total of 15 degrees F over a 5-day period. the temperature on the first day was 78 degrees F. What wa
Montano1993 [528]
I think it's 15 because it is
4 0
4 years ago
Please help me latterly i posted this and no one bothered too help me plz i beg of u i really need help u have no idea i am fail
Nostrana [21]

1. An equation for income, I = 3c + 2p, where c is the number of calendars sold and p is the number of posters sold.

2. When 25 calendars and 18 posters are sold, the income is $111.

3. Income for selling 12 calendars and 15 posters is $66.

4. 3 possible combinations; 1) 30 calendars and 5 posters. 2) 20 calendars and 20 posters. 3) 10 calendars and 35 posters.

Step-by-step explanation:

Step 1; It is given that each calendar costs $3 and each poster costs $2. So if c is the number of calendars sold and p is the number of posters sold.

Total income, I = 3c + 2p.

Step 2; Now if we substitute the value of c with 25 and p with 18, we get

I = 3c + 2p = 3 (25) + 2 (18) = 75 + 36 = $111.

Step 3; If the value of p is 12 and the value of c is 15, we have

I = 3c + 2p = 3 (12) + 2 (15) = 36 + 30 = $66.

Step 4; To find combinations that cost $100, we substitute I with 100. So the equation becomes 3c + 2p = 100.

When c = 30, 3 (30) + 2p = 100, 2p = 100 - 90 = 10, 2p = 10, p =5.

When c = 20, 3 (20) + 2p = 100, 2p = 100 - 60 = 40, 2p = 40, p = 20.

When c = 10, 3 (10) + 2p = 100, 2p = 100 - 30 = 70, 2p = 70, p = 35.

4 0
4 years ago
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