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4 years ago

5 children are playing on a trampoline.

Mathematics

1 answer:

Blizzard [7]4 years ago

4 0

Answer:

Step-by-step explanation:

The mean weight is the average

(x1+x2+x3+x4+x5)/5 = 28

Multiply by 5

(x1+x2+x3+x4+x5) =140

The total weight of the 5 children is 140

(x1+x2)/2 =26.5

Multiply by 2

(x1+x2)/2*2 =26.5*2

(x1+x2) =53

The total weight of the two children is 53

140-53 =87

The three children weigh 87

Divide by 3 to find the mean weight

(x3+x4+x5)/3 = 87/3

mean weight = 29

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There are 36 carpenters in a crew. On a certain day, 29 were present. What percent showed up for work.

PolarNik [594]

29/36=0.8055
0.8055*100
80.55% showed up for work

4 0

3 years ago

Amanda uses a rectangular canvas for a panting. The length is 6x-3 centimeters The width is 2x+6 centimeters,and is 4/5of the le

julia-pushkina [17]

Answer:

15 cm by 12 cm.

Step-by-step explanation:

Given:

Amanda uses a rectangular canvas for a panting.

The length is $6x-3$ centimeters.

The width is $2x+6$ centimeters, and is 4/5 of the length .

Question asked:

.What are the dimensions of the canvas ?

Solution:

As given that the width is $\frac{4}{5}$ of the length.

$2x+6=\frac{4}{5} (6x-3)\\ \\ 2x+6=\frac{4}{5}\times6x-\frac{4}{5}\times3\\ \\ 2x+6=\frac{24x}{5} -\frac{12}{5} \\ \\$

Adding both sides by $\frac{12}{5}$

$2x+6+\frac{12}{5} =\frac{24x}{5} -\frac{12}{5}+\frac{12}{5}\\ \\ 2x+\frac{42}{5} =\frac{24x}{5}$

Subtracting both sides by $2x$

$2x-2x+\frac{42}{5} =\frac{24x}{5} -2x\\ \\ \frac{42}{5} =\frac{24x-10x}{5} \\ \\ \frac{42}{5} =\frac{14x}{5} \\ \\$

By cross multiplication:-

$5\times14x=5\times42\\ \\ 70x=210\\ \\$

By diving both sides by 70

$x=3$

Now, substituting the value:-

The length = $6x-3$ cm

= $6\times3-3=18-3=15\ cm$

The width = $2x+6$ cm

= $2\times3+6=6+6=12\ cm$

Thus, length and width of canvas are 15 cm and 12 cm.

7 0

3 years ago

Can someone help me please ?

mafiozo [28]

25^(n-3)=5^(n+8)
Remember: 25=5²
Therefore:
5^[(n-3)²]=5^(n+8)
5^(2n-6)=5^(n+8)
Then:
2n-6=n+8
2n-n=8+6
n=14

Answer: n=14

7 0

3 years ago

Read 2 more answers

Nathan has 80 us stamps,64 Canadian stamps,and 32 Mexican stamps. He wants to put the same number of stamps on each page. Each p

Feliz [49]

Answer:

The greatest number of stamps that Nathan can put on each page = 16.

Step-by-step explanation:

Given:

Nathan has:

80 US stamps

64 Canadian stamps

32 Mexican stamps

The stamps need to put on a page such that each page has same number of same country stamps on each page.

To find the greatest number of stamps he can put on each page.

Solution:

In order to find the greatest number of stamps Nathan can put on each page, we will find the G.C.F. of the three numbers.

The numbers are:

$80,64,32$

We will list down the prime factors of each number.

$80=2\times 2\times 2\times 2\times 5$

$64=2\times 2\times 2\times 2\times 2\times 2$

$32=2\times 2\times2\times 2\times 2$

The G.C.F can be given as = $2\times 2\times 2\times 2$ = 16

Thus, the greatest number of stamps that Nathan can put on each page = 16.

6 0

3 years ago

Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x

djverab [1.8K]

I'll assume the ODE is

$y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}$

Solve the homogeneous ODE,

$y'' - 3y' + 2y = 0$

The characteristic equation

$r^2 - 3r + 2 = (r - 1) (r - 2) = 0$

has roots at $r=1$ and $r=2$ . Then the characteristic solution is

$y = C_1 e^x + C_2 e^{2x}$

For nonhomogeneous ODE (1),

$y'' - 3y' + 2y = e^x$

consider the ansatz particular solution

$y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x$

Substituting this into (1) gives

$a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1$

For the nonhomogeneous ODE (2),

$y'' - 3y' + 2y = e^{2x}$

take the ansatz

$y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}$

Substitute (2) into the ODE to get

$b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1$

Lastly, for the nonhomogeneous ODE (3)

$y'' - 3y' + 2y = e^{-x}$

take the ansatz

$y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}$

and solve for $c$ .

$ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16$

Then the general solution to the ODE is

$\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}$

6 0

1 year ago