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elena55 [62]
3 years ago
15

Select the equation written in point-slope form.

Mathematics
1 answer:
slavikrds [6]3 years ago
4 0
Poin slope form is
y-y1=m(x-1)
the answer is A
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It takes about 20 minutes to grade a student’s paper. How long, in hours, does it take to grade papers for a class of 30 people?
insens350 [35]

Answer:

The answer is 600 minutes (10 hours)

Step-by-step explanation:

20 minutes = 1 student.

30 students x 20 = 600 minutes

600mins / 60 (because an hour has 60 mins) = 10 hours.

5 0
3 years ago
Read 2 more answers
If 2/3(x-6)-3/4x=3 , then x= ?
Elan Coil [88]

Answer:

The value of x for the given expression is - 84

Step-by-step explanation:

Given expression as :

\frac{2}{3} (x - 6) - \frac{3}{4} x = 3

Taking LCM of 3 & 4

So, \frac{8\times (x-6) - 9x}{12} = 3

or, 8 x - 48 - 9 x = 12 × 3

Or, 8 x - 9 x = 36 + 48

or, - x = 84

∴  x = - 84

Hence The value of x for the given expression is - 84 Answer

4 0
3 years ago
Help meeee with this questionnnnnnnnnn!!!!!!!!!!!!
Dafna1 [17]

Given:

The scale factor between two circles is \dfrac{2x}{5y}.

To find:

The ratio of their areas.

Solution:

We know that, all circles are similar.

The ratio of the areas of similar figures is equal to the ratio of squares of their corresponding sides or equal to the square of ratio of their corresponding sides.

The scale factor is the ratio of the corresponding sides.

Ratio of areas of circles = Square of scale factor between two circles

\text{Ratio of areas of circles}=\left(\dfrac{2x}{5y}\right)^2

\text{Ratio of areas of circles}=\dfrac{(2x)^2}{(5y)^2}

\text{Ratio of areas of circles}=\dfrac{4x^2}{25y^2}

Therefore, the correct option is D.

3 0
3 years ago
Simplify by combining like terms x^2+2x+15-x-25
Romashka-Z-Leto [24]
X^2+x-10

does it help??
7 0
3 years ago
We would like to use the power series method to find the general solution to the differential equation d 2y dx2 − 4x dy dx + 12y
Feliz [49]

y=\displaystyle\sum_{n\ge0}a_nx^n

\dfrac{\mathrm dy}{\mathrm dx}=\displaystyle\sum_{n\ge1}na_nx^{n-1}\implies4x\dfrac{\mathrm dy}{\mathrm dx}=4\sum_{n\ge1}na_nx^n=4\sum_{n\ge0}na_nx^n

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n

Substituting into the ODE

\dfrac{\mathrm d^2y}{\mathrm dx^2}-4x\dfrac{\mathrm dy}{\mathrm dx}+12y=0

gives

\displaystyle\sum_{n\ge0}\bigg((n+2)(n+1)a_{n+2}-4na_n+12a_n\bigg)x^n=0

so that the coefficients of the series are given according to

\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_{n+2}=\dfrac{4(n-3)a_n}{(n+2)(n+1)}&\text{for }n\ge0\end{cases}

We can shift the index in the recursive part of this definition to get

a_n=\dfrac{4(n-5)a_{n-2}}{n(n-1)}

for n\ge2. There's dependency between coefficients that are 2 indices apart, so we can consider 2 cases:

  • If n=2k, where k\ge0 is an integer, then

k=0\implies n=0\implies a_0=a_0

but since y(0)=0, we have a_0=0 and a_{2k}=0 for all k\ge0.

  • If n=2k+1, then

k=0\implies n=1\implies a_1=a_1

k=1\implies n=3\implies a_3=\dfrac{4(-2)a_1}{3\cdot2}=-\dfrac43a_1

k=2\implies n=5\implies a_5=0

and so a_{2k+1}=0 for all k\ge2. If y'(0)=1, we then have a_1=1 and a_3=-\dfrac43.

So the ODE has solution

y(x)=\displaystyle\sum_{k\ge0}(a_{2k}x^{2k}+a_{2k+1}x^{2k+1})\implies\boxed{y(x)=x-\dfrac43x^3}

8 0
3 years ago
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