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Sloan [31]
2 years ago
11

For number 6 can anyone tell me the answer

Mathematics
1 answer:
slamgirl [31]2 years ago
7 0
The perimeter of MATH is 4w + 6 and the perimeter of GEOM is 6w - 8. the perimeters are the same so set them equal,
  4w + 6 = 6w - 8... use inverse to get all like terms on opposite sides
-4w + 8 = -4w + 8
        14 = 2w... divide both side by 2 to get w alone
       ÷2  = ÷2
          7 = w
so B. 7 is the answer
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What is the n in 4 1/2 = 1 1/2n; n =
jasenka [17]

Answer:

9/11

Step-by-step explanation:

4 1/2= 9/2=11/2n

so n =9/11

so 9/2=11/2*9/11

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2 years ago
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GalinKa [24]

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8 0
3 years ago
Need help fast please! Factor by grouping. 6x³ + 24x² -7x -28 =
olchik [2.2K]

Answer:

(x+4)(6x^2 - 7)

Step-by-step explanation:

Focus on the first 2 terms first and on the second 2 terms last:

6x^3 + 24x^2 = 6x^2(x+4)

-7x-28 = -7(x+4)

We see that the factor (x+4) is common to both pairs:  common to the first 2 terms and common to the last 2 terms.

Thus,

6x³ + 24x² -7x -28 = (x+4)(6x^2 - 7(x+4)

Factoring out x+4, we get (6x^2 - 7), and so 6x³ + 24x² -7x -28 in factored form is (x+4)(6x^2 - 7).

5 0
3 years ago
Expand the expression using the distributive property: 6(4x + 5y + 2)
Oksanka [162]

Answer:

24 x + 30 y + 12

Step-by-step explanation:

this is simplified

5 0
3 years ago
A tank initially has 300 gallons of a solution that contains 50 lb. of dissolved salt. A brine solution with a concentration of
belka [17]

Let <em>s(t)</em> be the amount of salt in the tank at time <em>t</em>. Then <em>s(0)</em> = 50 lb.

Salt flows into the tank at a rate of

(2 gal/min) (6 lb/gal) = 12 lb/min

and flows out at a rate of

(2 gal/min) (<em>s(t)</em>/300 lb/gal) = <em>s(t)</em>/150 lb/min

so that the net rate at which salt is exchanged through the tank is

d<em>s(t)</em>/d<em>t</em> = 12 - <em>s(t)</em>/150 … (lb/min)

Solve for <em>s(t)</em>. The DE is separable, so we have

d<em>s</em>/d<em>t</em> = 12 - <em>s</em>/150

150 d<em>s</em>/d<em>t</em> = 1800 - <em>s</em>

150/(1800 - <em>s</em>) d<em>s</em> = d<em>t</em>

Integrate both sides to get

-150 ln|1800 - <em>s</em>| = <em>t</em> + <em>C</em>

Solve for <em>s</em> :

ln|1800 - <em>s</em>| = -<em>t</em>/150 + <em>C</em>

1800 - <em>s</em> = exp(-<em>t</em>/150 + <em>C </em>)

1800 - <em>s</em> = <em>C</em> exp(-<em>t</em>/150)

<em>s</em> = 1800 - <em>C</em> exp(-<em>t</em>/150)

Now given that <em>s(0)</em> = 50, we solve for <em>C</em> :

50 = 1800 - <em>C</em> exp(-0/150)

50 = 1800 - <em>C</em>

<em>C</em> = 1750

Then the amount of salt in the tank at any time <em>t</em> ≥ 0 is

<em>s(t)</em> = 1800 - 1750 exp(-<em>t</em>/150)

To find the time it takes for the tank to hold 100 lb of salt, solve for <em>t</em> in

100 = 1800 - 1750 exp(-<em>t</em>/150)

1700 = 1750 exp(-<em>t</em>/150)

34/35 = exp(-<em>t</em>/150)

ln(34/35) = -<em>t</em>/150

<em>t</em> = -150 ln(34/35) ≈ 4.348

So it would take approximately 4.348 minutes.

By the way, we didn't have to solve for <em>s(t)</em>, we could have instead stopped with

-150 ln|1800 - <em>s</em>| = <em>t</em> + <em>C</em>

Solve for <em>C</em> - this <em>C</em> <u>is not</u> the same as the one we found using the other method. <em>s(0)</em> = 50, so

-150 ln|1800 - 50| = 0 + <em>C</em>

<em>C</em> = -150 ln|1750|

==>   <em>t</em> = 150 ln(1750) - 150 ln|1800 - <em>s</em>|

Then <em>s(t)</em> = 100 lb when

<em>t</em> = 150 ln(1750) - 150 ln(1700)

<em>t</em> = 150 ln(1750/1700)

<em>t</em> = 150 ln(35/34)

<em>t</em> ≈ 4.348

5 0
3 years ago
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