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3 years ago

A student found the slope of a line that passes through the points (-3, -10) and (5, -7) to be . What mistake did he make?

Mathematics

1 answer:

DedPeter [7]3 years ago

6 0

3/8 would be the correct slope, so D. He did not make a mistake. That is the correct slope.

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What number is 7 units to the left of -1? 8 6 -8 -6

Viefleur [7K]

Answer:

7 units to the left is -6

Step-by-step explanation:

your counting back wards but its not a neg. its positive so the answer is 6

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2 years ago

Solve. X^2+5x-10=0 Enter your answers, as exact values, in the boxes.

Grace [21]

Answer:

$x^{2} + 5x - 10 = 0 \\x^{2} + 5x = 10 \\ x(x + 5) = 10$

Now,

Either,

x=10

Or,

x+5=10

x=10-5

x=5

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2 years ago

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2 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

1 year ago

What is the lcm of 3,9, and 15?

s344n2d4d5 [400]

The least common multiple that is common in all numbers but is the lowest is 3 and it's the only number that all numbers listed can be divided easily by.

Hope this helps!

3 0

3 years ago

Read 2 more answers