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liubo4ka [24]
3 years ago
15

a class collects 54 pounds of food each holiday food drive how many 6 pound bags can the class collect

Mathematics
1 answer:
k0ka [10]3 years ago
8 0
54/6=9 6pound bags the class can collect.
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The question is Stephen read for 34 minutes on Monday and 45 minutes on Tuesday. Use breaking apart to find how many minutes Ste
user100 [1]
Breaking apart means that you just take the tens from each number, and the ones from each number, then add them up.


The tens from your numbers are:
30 from 34
40 from 45

The ones from your numbers are:
4 from 34
5 from 45

Now, you have to add up the tens and ones separately.
30 + 40 = 70
4 + 5 = 9

Finally, you add up those 2 numbers that you got- 70 and 9

The answer: Stephen read for 79 minutes total.
7 0
3 years ago
Line FG contains points F (3,7) and G (-4,-5) line HI contains points (-1,0) and I (4,6) lines FG and HI are. ?
ludmilkaskok [199]
FG : (3,7)(-4,-5)
slope = (-5 - 7) / (-4-3) = -12/-7 = 12/7

y = mx + b
slope(m) = 12/7
(3,7)...x = 3 and y = 7
now we sub, we r looking for b, the y int
7 = 12/7(3) + b
7 = 36/7 + b
7- 36/7 = b
49/7 - 36/7 = b
13/7 = b
so ur equation is : y = 12/7 + 13/7.....slope = 12/7, y int = 13/7

HI : (-1,0)(4,6)
slope = (6 - 0) / (4 - (-1) = 6/5

no need to go any farther.....these lines have different slopes...and their not negative reciprocals....so there will be one solution. Answer is : neither.
7 0
3 years ago
Read 2 more answers
The graph of
ratelena [41]

Answer:

Positive and negative intervals on a graph

Step-by-step explanation:

The positive regions of a function are those intervals where the function is above the x-axis. It is where the y-values are positive (not zero). The negative regions of a function are those intervals where the function is below the x-axis. It is where the y-values are negative (not zero). hope this helps you :)

3 0
3 years ago
Read 2 more answers
CALCULUS: Determine which function is a solution to the differential equation y ' − y = 0.
Montano1993 [528]

C: none of these are solutions to the given equation.

• If<em> y(x)</em> = <em>e</em>², then <em>y</em> is constant and <em>y'</em> = 0. Then <em>y'</em> - <em>y</em> = -<em>e</em>² ≠ 0.

• If <em>y(x)</em> = <em>x</em>, then <em>y'</em> = 1, but <em>y'</em> - <em>y</em> = 1 - <em>x</em> ≠ 0.

The actual solution is easy to find, since this equation is separable.

<em>y'</em> - <em>y</em> = 0

d<em>y</em>/d<em>x</em> = <em>y</em>

d<em>y</em>/<em>y</em> = d<em>x</em>

∫ d<em>y</em>/<em>y</em> = ∫ d<em>x</em>

ln|<em>y</em>| = <em>x</em> + <em>C</em>

<em>y</em> = exp(<em>x</em> + <em>C </em>)

<em>y</em> = <em>C</em> exp(<em>x</em>) = <em>C</em> <em>eˣ</em>

8 0
3 years ago
Hello, please help thank youu!!
uranmaximum [27]

Answer:

36 \frac{1}{2}

Step-by-step explanation:

<u>(5</u>\frac{1}{2}<u> x 3)</u> + (10 ÷ \frac{1}{2})

(16\frac{1}{2}) + <u>(10 ÷ </u>\frac{1}{2}<u>)</u>

<u>(16</u>\frac{1}{2}<u>) + (20)</u>

36 \frac{1}{2}

7 0
3 years ago
Read 2 more answers
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