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crimeas [40]
2 years ago
7

Help mmmmeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

Mathematics
2 answers:
kaheart [24]2 years ago
8 0

Answer:

What help you need from us?????

Lisa [10]2 years ago
7 0

Answer:

I think the answer is 6

Step-by-step explanation:

but it could also be 5 since its the lowest number. But if its goin based of the amount of data it should be 6

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A 5x5 table is filled with numbers (one number per cell). It is known that the product of 5 numbers in each column is negative.
Advocard [28]

Answer:

Yes

Step-by-step explanation:

A negative multiplied by another negative will turn into a positive. A positive multipled by a negative will stay as a negative.

5 0
2 years ago
Find the product. x3(x2+5x+1)
PtichkaEL [24]

Answer:

Answer = x5+5x4+x3

Step-by-step explanation:

x3(x2+5x+1)

=(x3)(x2+5x+1)

=(x3)(x2)+(x3)(5x)+(x3)(1)

=x5+5x4+x3

Answer = x5+5x4+x3

7 0
2 years ago
Solve 6x^2 = 12x -24
Mice21 [21]

Answer:

number 2)

Explanation step by step

7 0
3 years ago
Read 2 more answers
4(3y – 6) = -3(8 – 4y)
Nadya [2.5K]

Answer:

0=0

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

4(3y−6)=−3(8−4y)

(4)(3y)+(4)(−6)=(−3)(8)+(−3)(−4y)(Distribute)

12y+−24=−24+12y

12y−24=12y−24

Step 2: Subtract 12y from both sides.

12y−24−12y=12y−24−12y

−24=−24

Step 3: Add 24 to both sides.

−24+24=−24+24

0=0

3 0
3 years ago
Andrew has a cell phone plan that provides 300 free minutes each month for a flat rate of $19. for any minutes over 200, andrew
natima [27]

Answer:

f (x) = { 19 + 0.39(x - 300), x > 300

Step-by-step explanation:

Andrew has a cell phone plan that provides 300 free minutes each month for a flat rate of $19. For any minutes over 300, Andrew is charged $0.39 per minute. Let x be the number of minutes Andrew uses per month and f(x) be the charges based on Andrew's cell phone plan. If  then If  then first 300 minutes are free and each minute of next (x-300) minutes costs $0.39, therefore Hence, { 19 + 0.39(x - 300), x > 300

Hoped I helped

7 0
3 years ago
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