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brilliants [131]
3 years ago
9

Which is a common denominator for the numbers: 40 and 28

Mathematics
1 answer:
Studentka2010 [4]3 years ago
4 0

Hi!

I can help you with joy!

Let's list 10 multiples of 40 and 10 multiples of 28.

The first 10 multiples of 40 are:

40, 80, 120, 160, 200, 240, 280, 320, 360, 400...

The first 10 multiples of 28 are:

28, 56, 84,112, 140, 168, 196, 224, 252, 280

I have highlighted it. Guess why!

Because it's the LCM (LEAST COMMON MULTIPLE) of 40 and 48.

*RosalindCordelia*

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Why is the initial value and y-intercept the same thing on a graph?? someone please explain
lapo4ka [179]

-y is the vertical line and -x is the horizontal line. The only reason they would say -y 3 is to let you know that is the vertical... so you don’t mess up and mark 3 on the -x (horizontal) line

8 0
3 years ago
Ann takes 70 paces to walk 50m. the number of paces ann takes to walk 3.5km is A.70 B.490. C.3900 D.4900​
Shalnov [3]
<h3>Answer:  4900  (choice D)</h3>

===================================================

Work Shown:

1 km = 1000 m

3.5 km = 3500 m  ............. multiplying both sides by 3.5

(70 paces)/(50 m) = (x paces)/(3500 m)

70/50 = x/3500

7/5 = x/3500

7*3500 = 5x .............. cross multiply

5x = 7*3500

x = (7*3500)/5

x = (7*5*700)/5

x = 7*700

x = 4900

Ann takes 4900 paces to walk 3.5 km

4 0
2 years ago
Complete the inequality statement <br> 4/13 __6/13
nydimaria [60]
< answer because 4/13 is less than
5 0
3 years ago
A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of
saw5 [17]

Let A(t) denote the amount of salt in the tank at time t.

Salt flows in at a rate of

(1 lb/gal) * (3 gal/min) = 3 lb/min

and flows out at a rate of

(A(t)/(200 + t) lb/gal) * (2 gal/min) = 2 A(t)/(500 + t)

(in case you're unsure about the denominator: the tank starts off with 200 gal of solution, and each minute solution flows in at a rate of 3 gal/min and thus the tank gains (3 gal/min) * (1 min) = 3 gal. At the same time, solution flows out at a rate of 2 gal/min and thus the tank loses 2 gal, giving a net change in volume of (3 - 2)*t = t gal)

Then the net rate of salt flow is given by the ODE,

\dfrac{\mathrm dA(t)}{\mathrm dt}-\dfrac{2A(t)}{200+t}=3

Multiply both sides by (200+t)^{-2}:

(200+t)^{-1}\dfrac{\mathrm dA(t)}{\mathrm dt}-2(200+t)^{-3}A()=3(200+t)^{-2}

\implies\dfrac{\mathrm d}{\mathrm dt}\bigg((200+t)^{-2}A(t)\bigg)=3(200+t)^{-2}

Integrating both sides and solving for A(t) gives

(200+t)^{-2}A(t)=-\dfrac3{200+t}+C

A(t)=-2(200+t)+C(200+t)^2

The tank starts off with 100 lb of salt in solution, so A(0)=100 and we find

100=-2(200)+C(200)^2\implies C=\dfrac1{80}

and so

A(t)=-2(200+t)+\dfrac{(200+t)^2}{80}=\dfrac{(200+t)(40+t)}{80}

The tank will begin to overflow once the volume of solution reaches 500 gal; this happens when

500=200+t\implies t=300

or 300 minutes or 5 hours after solution starts flowing. At this point, the tank will contain

A(300)=2125

or 2125 lb of salt.

Theoretically, the amount of salt in the tank will increase forever, since A(t)\to\infty as t\to\infty.

6 0
4 years ago
Pls help with these questions!!!
DedPeter [7]

Answer:

9 \times 9 - 3 = 81 - 3 = 78

6 0
3 years ago
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