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egoroff_w [7]
3 years ago
11

Greg needs at least $1.60 in stamps to mail a package. He has 28cent stamps and 4cent stamps. He can use no more than twenty 4ce

nt stamps as he only has one book left.
Write a system of linear inequalities to represent this situation.
Mathematics
1 answer:
lapo4ka [179]3 years ago
7 0

19 4-cent stamps and 3 28-cent stamps should be used.

Given that Greg needs at least $ 1.60 in stamps to mail a package, and he has 28cent stamps and 4cent stamps, and he can use no more than twenty 4cent stamps as he only has one book left, to determine how many stamps to use should be made the following calculation, through a linear function:

  • (0.04 x 20) + 0.28X = 1.60
  • 0.80 + 0.28X = 1.60
  • 0.28X = 1.60 - 0.80
  • 0.28X = 0.80
  • X = 0.8 / 0.28
  • X = 2.85

  • (3 x 0.28) + 0.04X = 1.60
  • 0.84 + 0.04X = 1.60
  • 0.04X = 1.60 - 0.84
  • 0.04X = 0.76
  • X = 0.76 / 0.04
  • X = 19

Therefore, 19 4-cent stamps and 3 28-cent stamps should be used.

Learn more in brainly.com/question/15325318

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Calculate the slant height for the given square pyramid. Round to the nearest tenth.
spayn [35]

the answer would have to be 7.8


7 0
3 years ago
Can you solve the system of equation using substitution y=x+8 <br> X+y=2
Vsevolod [243]
X + y = 2
Substitute x + 8 into y
x + x + 8 = 2
2x + 8 =2
(Subtract 8 from both sides)
2x = -6
(Divide both sides by 2)
x = -3

Subsitute x = -3 into either equation to find y
x + y = 2
-3 + y = 2
(Add 3 to both sides)
y = 2 + 3
y = 5

Or y can be solved for using:
y = x + 8
y = -3 + 8
y = 5
8 0
3 years ago
Find all the solutions for the equation:
Contact [7]

2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0

Divide both sides by x^2\,\mathrm dx to get

2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}

Substitute v(x)=\dfrac{y(x)}x, so that \dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}. Then

x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}

x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}

x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}

The remaining ODE is separable. Separating the variables gives

\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x

Integrate both sides. On the left, split up the integrand into partial fractions.

\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}

\implies v^2+2v+1=a(v^2+1)+(bv+c)v

\implies v^2+2v+1=(a+b)v^2+cv+a

\implies a=1,b=0,c=2

Then

\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v

On the right, we have

\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C

Solving for v(x) explicitly is unlikely to succeed, so we leave the solution in implicit form,

\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C

and finally solve in terms of y(x) by replacing v(x)=\dfrac{y(x)}x:

\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C

\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C

\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}

7 0
3 years ago
Urgent !!!!!!!!!!!!!!! 10 points
Mademuasel [1]

Answer:

136 cm²

Step-by-step explanation:

Surface area = 2(lw+wh+hl)

l = 7

w = 2

h = 6

so,

2(7×2+2×6+7×6)

= 136 cm²

6 0
2 years ago
Read 2 more answers
Find the median of the set of data. 30, 16, 49, 25 <br><br>​
weeeeeb [17]

Answer:

16,25,30,49

25+30=55

55÷2=27.5

Step-by-step explanation:

median means the middle value of any set of data so first arrange the data into ascending order.

16, 25, 30, 49

the data is even so we take both the middle value , add it and divide it with 2

25+30=55

55÷2=27.5

4 0
3 years ago
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