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Gre4nikov [31]
2 years ago
13

Work out the equation of the line which has a gradient of 2 and passes through the point (1,4)

Mathematics
1 answer:
Vladimir79 [104]2 years ago
4 0

Answer:

see photo for detailed analysis

Step-by-step explanation:

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Write an expression for calculation double the product of 4 and 7
Over [174]

Answer:2(4*7)

this is an expression

Step-by-step explanation:

if you want to solve it  

you would multiply 4 by 7 which gives you 28

then double 28

you would multiply 28 by 2

which then gives you an answer of 56

3 0
3 years ago
What are the solutions of the equation 9x^4 – 2x^2 – 7 = 0? Use u substitution to solve.
Debora [2.8K]
9x^4 - 2x^2 - 7 = 0 \\ \text{substitution: }  x^2=t, \ \ t\ \textgreater \ 0\\ 9t^2-2t-7\ \textgreater \ 0 \\ D=b^2-4ac=(-2)^2-4*9*(-7)=4+252 = 256 \\ t_{1,2}= \frac{-bб \sqrt{D} }{2a}= \frac{2б 16 }{18} \\ t_1=- \frac{14}{18} \ \ \ \ \O \\  \boxed{t_2=1} \\ \\x^2=1\\x=б \sqrt{1} \\ x=б1

Answer: x=-1, x=1
8 0
3 years ago
anne buys used cars at auction for 2000 per car. there is 150 fee to take part in the auction. If anne buys 13 used cars how muc
katrin [286]
How to understand the problem is like this:
   Total Cost = Entrance Fee + Total Car cost
So we know that the Entrance Fee is $150...
   Total Cost = $150 + Total Car cost
So we know that Anne bought 13 cars, each at $2,000...
   Total Cost = $150 + ($2,000 x 13)
We multiply because each car is $2,000 and Anne bought 13 of them.
So it works out like this:
   Total Cost = $150 + ($2,000 x 13)
   Total Cost = $150 + (26000)
   Total Cost = $26150
5 0
4 years ago
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}}

4 0
2 years ago
Difference of two exponential random variables
torisob [31]
The difference would be subtracting the variables and the dividing the difference by 2

8 0
3 years ago
Read 2 more answers
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