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xeze [42]
2 years ago
15

The sum of all three numbers is 12. The sum of twice the first number, 4 times the second number, and 5 times the third number i

s 35. The difference between 8 times the first number and the second number is 52. What are the three numbers?
Mathematics
1 answer:
DaniilM [7]2 years ago
4 0

Answer:

The numbers are 7, 4 and 1

Step-by-step explanation:

Let the numbers =x, y, z.

The sum of all three numbers is 12, x+y+z=12--------i

The sum of twice the first number, 4 times the second number, and 5 times the third number is 35, 2x+4y+5z=35-------ii

The difference between 8 times the first number and the second number is 52, 8x-y=52---------iii

Now, from (i),z =1-x-y--------iv

substitute (iv) in (ii),

2x+4y+5(12-x-y)=35

2x+4y+60-5x-5y=35

collect like terms

-3x-y=-25

that is, 3x-y=25--------v

solving (iii) and (v) simultaneously (add iii and v), we have

11x=77

divide both sides by 11 to obtain x=7

Put x=7 in (v)

3(7)+y=25

21+y=25

implies y=25-21=4

put x=7 and y=4 in (iv)

z=12-7-4=1

Hence, the numbers are 7, 4 and 1.

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2 years ago
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Zepler [3.9K]

Answer:

178.3 mm²

Step-by-step explanation:

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Sₐ = 43.3 + 3 x (1/2 x (10 x 9)) = 43.3 + (3 x 45) = 178.3

5 0
3 years ago
adults say that they have cheated on a test or exam before. You randomly select six adults. Find the probability that the number
VLD [36.1K]

Answer: a. 1/6. b. 2/3. c. 5/6

Step-by-step explanation:

Number of adults randomly selected= 6

a. Finding the probability of adults that have cheated on an examination before is exactly 4.

The numbers are 1, 2, 3, 4, 5 and 6.

Probability of chosen exactly 4 will be 1/6 since 4 can only occur once.

b. More than 2means 3,4, 5 and 6.

Probability that the number of adults that have cheated on a test or exam is more than 2 will be:

4/6 = 2/3

c. At most 5 means 1,2,3,4 or 5

This will give 5/6

5 0
3 years ago
Question 2 &lt;&gt;&lt;&gt;&lt;&gt;&lt;&gt;&lt;&gt;&lt;&gt;&lt;&gt;&lt;&gt;&lt;&gt;&lt;&gt;&lt;&gt;
drek231 [11]

Answer:

I deserve brainliest just as all good boys deserves fudgeeeeee

Step-by-step explanation:

v^2 = 25/81

v= the square root of 25/81 = 5/9

since that isn't a choise, don't simplify.

D works too  because negative x negative = positive

CCCCCCC

4 0
3 years ago
Read 2 more answers
If x^2y-3x=y^3-3, then at the point (-1,2), (dy/dx)?
zavuch27 [327]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2866883

_______________


          dy
Find  ——  for an implicit function:
          dx


x²y – 3x = y³ – 3


First, differentiate implicitly both sides with respect to x. Keep in mind that y is not just a variable, but it is also a function of x, so you have to use the chain rule there:

\mathsf{\dfrac{d}{dx}(x^2 y-3x)=\dfrac{d}{dx}(y^3-3)}\\\\\\&#10;\mathsf{\dfrac{d}{dx}(x^2 y)-3\,\dfrac{d}{dx}(x)=\dfrac{d}{dx}(y^3)-\dfrac{d}{dx}(3)}


Applying the product rule for the first term at the left-hand side:

\mathsf{\left[\dfrac{d}{dx}(x^2)\cdot y+x^2\cdot \dfrac{d}{dx}(y)\right]-3\cdot 1=3y^2\cdot \dfrac{dy}{dx}-0}\\\\\\&#10;\mathsf{\left[2x\cdot y+x^2\cdot \dfrac{dy}{dx}\right]-3=3y^2\cdot \dfrac{dy}{dx}}


                        dy
Now, isolate  ——  in the equation above:
                        dx

\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3=3y^2\cdot \dfrac{dy}{dx}}\\\\\\&#10;\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3-3y^2\cdot \dfrac{dy}{dx}=0}\\\\\\&#10;\mathsf{x^2\cdot \dfrac{dy}{dx}-3y^2\cdot \dfrac{dy}{dx}=-\,2xy+3}\\\\\\&#10;\mathsf{(x^2-3y^2)\cdot \dfrac{dy}{dx}=-\,2xy+3}


\mathsf{\dfrac{dy}{dx}=\dfrac{-\,2xy+3}{x^2-3y^2}\qquad\quad for~~x^2-3y^2\ne 0}


Compute the derivative value at the point (– 1, 2):

x = – 1   and   y = 2


\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{-\,2\cdot (-1)\cdot 2+3}{(-1)^2-3\cdot 2^2}}\\\\\\&#10;\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{4+3}{1-12}}\\\\\\&#10;\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{7}{-11}}\\\\\\\\ \therefore~~\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=-\,\dfrac{7}{11}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)


Tags:  <em>implicit function derivative implicit differentiation chain product rule differential integral calculus</em>

6 0
2 years ago
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