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Roman55 [17]
3 years ago
14

One number is 33 times a first number. a third number is 100 more than the first number. if the sum of the three numbers is 3103

10​, find the numbers.
Mathematics
1 answer:
kherson [118]3 years ago
7 0
Create some expressions.

1# number: x

2# number: 33x

3# number: 100 + x


this all equals to 310310 when added so

x+33x+(x+100) = 310310

Solve what you know first then rearrange it.

34x + x + 100 = 310310

35x = 310310-100

35x = 310210

x = 310210/35

x = 8863.142857

Substitute the x into the expressions made and you'll find the 2# and 3# number. You've already found the 1# number since it's only x.
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Use slope-intercept form to write the equation of a line
Korvikt [17]

Answer:

y=-3x-2

Step-by-step explanation:

There is enough information to make a point-slope form equation that which we can convert into slope-intercept form.

Point-slope form is: y-y_1=m(x-x_1)

We are given the slope of -3 and the point of (1,-5).

y-y_1=m(x-x_1)\rightarrow y+5=-3(x-1)

Convert into Slope-Intercept Form:

y+5=-3(x-1)\\y+5-5=-3(x-1)-5\\\boxed{y=-3x-2}

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2 years ago
What is the sum of 1/8+5/16+3/8
Murljashka [212]

Answer:

i think it's 13/16

Step-by-step explanation:

1/8 times 2 is 2/16 and 3/8 times 2 is 6/16 so add all up to get 13/16

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2 years ago
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Please help me with the below question.
VMariaS [17]

By letting

y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}

we get derivatives

y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}

y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0

Examine the lowest degree term \left(x^{r-1}\right), which gives rise to the indicial equation,

5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients c_k is

(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}

so that with r = 4/5, the coefficients are governed by

c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}

c) Starting with c_0=1, we find

c_1 = -\dfrac{c_0}5 = -\dfrac15

c_2 = -\dfrac{c_1}{10} = \dfrac1{50}

so that the first three terms of the solution are

\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}

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2 years ago
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stira [4]

Step-by-step explanation:

Let's get our relevant base equations listed out:

d = 2r

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Thus, d=2*5m  = 10m, c=π*10m=10π m, and area = 25m²*π

Feel free to ask further questions!

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Answer:

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Step-by-step explanation:

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