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zaharov [31]
3 years ago
9

Find the missing factor 7s^2+25s+12=(s+3)()

Mathematics
1 answer:
Deffense [45]3 years ago
8 0
7s^2+25s+12=\\
7s^2+21s+4s+12=\\
7s(s+3)+4(s+3)=\\
(s+3)(7s+4)

It's 7s+4.
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Please help ASAP
Arisa [49]

Answer:

2 lessons per day for 27 days and 3 lessons a day for 4 days.

Step-by-step explanation:

66/31 = 2 4/31

Since 66 divided by 31 is 2 with remainder 4, you need to do 31 days of 2 lessons per day. To do the extra 4 lessons, do one more lesson per day for 4 of those days.

That results in:

2 lessons per day for 27 days and 3 lessons a day for 4 days.

Check:

2 lessons a day for 27 days = 2 * 27 = 54

3 lessons a day for 4 days = 3 * 4 = 12

54 + 12 = 66

If you do 2 lessons a day for 27 days and 3 lessons a day for 4 days, you will do the 66 lessons.

5 0
3 years ago
3 (x + 1) - 5 = 3x - 2<br> has how many solutions?
marissa [1.9K]

Answer:

Zero Solutions

Step-by-step explanation:

3x+3-5=3x-2

3x-3x= -2+5

0= 3

0 does not equal 3, so zero solutions.

7 0
3 years ago
Evaluate each expression<br><br> g5-h3 if g=2 and h=7
vovangra [49]
Look at it this way:
g=2
H=7
so it would be 2*5-7*3
than just do the multiplication first so 2*5=10 and 7*3=21 than subtract the two 10-21=-11
7 0
3 years ago
Please help me. Show work
Goryan [66]
What hell there is nothings ??
7 0
3 years ago
3. Find two possible lengths for CD if C, D, and E
Kaylis [27]

Given :

C, D, and E  are col-linear, CE = 15.8 centimetres, and DE=  3.5 centimetres.

To Find :

Two possible lengths for CD.

Solution :

Their are two cases :

1)

When D is in between C and E .

.                   .           .

C                  D          E

Here, CD = CE - DE

CD = 15.8 - 3.5 cm

CD = 12.3 cm

2)

When E is in between D and C.

.       .                .

D      E              C

Here, CD = CE + DE

CD = 15.8 + 3.5 cm

CD = 19.3 cm

Hence, this is the required solution.

3 0
3 years ago
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