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ElenaW [278]
3 years ago
15

A BOOKMARK

Mathematics
1 answer:
Anna [14]3 years ago
5 0

Answer:

The answer is 3/4

Step-by-step explanation:

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Help me solve this equation and show work <br><br> 11d + 17b = 95 <br> d + b = 7
sweet-ann [11.9K]

Answer:

{d,b}={4,3}

Step-by-step explanation:

[1]    11d + 17b = 95

  [2]    d + b = 7

Graphic Representation of the Equations :

   17b + 11d = 95        b + d = 7  

 

Solve by Substitution :

// Solve equation [2] for the variable  b

 [2]    b = -d + 7

// Plug this in for variable  b  in equation [1]

  [1]    11d + 17•(-d +7) = 95

  [1]    -6d = -24

// Solve equation [1] for the variable  d

  [1]    6d = 24

  [1]    d = 4

// By now we know this much :

   d = 4

   b = -d+7

// Use the  d  value to solve for  b

   b = -(4)+7 = 3

Solution :

{d,b} = {4,3}

4 0
3 years ago
Read 2 more answers
What is the average of 69 and 50?<br>PLEASE HELP<br>​
hjlf
(69+50)/2
(119)/2
59.5
7 0
3 years ago
Does 6(x + 5) = 6x + 11 , have on solution, infinitely many solutions, or no solutions?
choli [55]

Answer:

no solution

Step-by-step explanation:

hello

6(x+5)=6x+11\\ 6x+30=6x+11\\ 30=11

this is always false so there is no solution

3 0
3 years ago
Read 2 more answers
liam walked 100m around the perimeter of a rectangular playground. How long and how wide could the playground be? give 3 differe
Zolol [24]
You could do 30,30,20,20
Or 40,40,10,10
Or 45,45,5,5
4 0
3 years ago
Bernoulli differential equation... y'+xy=xy^2
snow_lady [41]
y'+xy=xy^2\implies y^{-2}y'+xy^{-1}=x

Let z=y^{-1}, so that z'=-y^{-2}y'. Then the ODE becomes linear in z with

-z'+xz=x\implies z'-xz=-x

Find an integrating factor:

\mu(x)=\exp\left(\displaystyle\int-x\,\mathrm dx\right)=e^{-x^2/2}

Multiply both sides of the ODE by \mu:

e^{-x^2/2}z'-xe^{-x^2/2}z=-xe^{-x^2/2}

The left side can be consolidated as a derivative:

\left(e^{-x^2/2}z\right)'=-xe^{-x^2/2}

Integrate both sides with respect to x to get

e^{-x^2/2}z=e^{x^2/2}+C

where the right side can be computed with a simple substitution. Then

z=1+Ce^{x^2/2}

Back-substitute to solve for y.

y^{-1}=1+Ce^{x^2/2}\implies y=\dfrac1{1+Ce^{x^2/2}}
3 0
3 years ago
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