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sasho [114]
2 years ago
8

Solve the equation for d. 0.2(d-6)=0.3d+5-3+0.1d

Mathematics
2 answers:
maria [59]2 years ago
7 0

Answer:

Step-by-step explanation:

0.2(d - 6) = 0.3d + 5 - 3 + 0.1d

0.2d - 1.2 = 0.4d + 2

     - 0.2d = 3.2

           d = - 16

Vsevolod [243]2 years ago
5 0

Answer:

d = -16

Step-by-step explanation:

0.2(d-6)=0.3d+5-3+0.1d\\\rule{150}{0.5}\\0.2d - 1.2 = 0.3d+5-3+0.1d\\\\0.2d - 1.2 = 0.3d+0.1d+ 5-3\\\\0.2d - 1.2=0.4d + 2\\\\0.2d- 0.2d - 1.2 = 0.4d -0.2d + 2\\\\-1.2 = 0.2d + 2\\\\-1.2 -2 = 0.2d + 2 - 2\\\\-3.2 = 0.2d\\\\\frac{-3.2 = 0.2d}{0.2}\\\\-16 = d\\\\\boxed{d = -16}

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The length of a rectangle is 4 yd longer than its width. If the perimeter of the rectangle is 48 yd, find its area.
bagirrra123 [75]
48 = 2(w+4) + 2w
48 = 4w+8
40 = 4w
w= 10 yd
A= 10 * 14
A = 140 yd
7 0
2 years ago
The doubling period of a bacterial population is 15 minutes. At time t= 80 minutes, the bacterial population was 90000​
Papessa [141]

Answer:

here i finished!

hope it helps yw!

Step-by-step explanation:

The doubling period of a bacterial population is 15 minutes.

At time t = 90 minutes, the bacterial population was 50000.

Round your answers to at least 1 decimal place.

:

We can use the formula:

A = Ao*2^(t/d); where:

A = amt after t time

Ao = initial amt (t=0)

t = time period in question

d = doubling time of substance

In our problem

d = 15 min

t = 90 min

A = 50000

What was the initial population at time t = 0

Ao * 2^(90/15) = 50000

Ao * 2^6 = 50000

We know 2^6 = 64

64(Ao) = 50000

Ao = 50000/64

Ao = 781.25 is the initial population

:

Find the size of the bacterial population after 4 hours

Change 4 hr to 240 min

A = 781.25 * 2^(240/15

A = 781.25 * 2^16

A= 781.25 * 65536

A = 51,199,218.75 after 4 hrs

6 0
3 years ago
X = A 21.6 B 12.9 C 9.1
Artist 52 [7]

Answer:

B: 12.9

Step-by-step explanation:

If A B and C are the alternatives, the answer is B: 12.9

The reason is that the hypotenuse can't be smaller than one of its sides, that eliminates alternative C.

And also, the hypotenuse can't be bigger than segments AC + BC, since the side AC is bigger and AC measures 10.8, the hypotenuse would have to measure < 21.6

Any questions, comment.

6 0
2 years ago
Read 2 more answers
PLEASE HELP
kifflom [539]

\\ \sf\longmapsto \dfrac{x^2-1}{x^2+5x+4}\leqslant 0

  • Solving denominator

\\ \sf\longmapsto x^2+5x+4>0

\\ \sf\longmapsto x^2+4x+x+4>0

\\ \sf\longmapsto x(x+4)+1(x+4)>0

\\ \sf\longmapsto (x+4)(x+1)>0

\\ \sf\longmapsto x>-4\:or x>-1

  • Hence x\neq-1
  • x can't be 0 as it makes function undefined

\\ \sf\longmapsto -4

5 0
2 years ago
H=4(x+3y)+2 make x the subject
blagie [28]
H = 4(x + 3y) + 2
H = 4x + 12y + 2
H - 12y - 2 = 4x
(H - 12y - 2) / 4 = x or 1/4H - 3y - 1/2 = x

5 0
2 years ago
Read 2 more answers
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