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Finger [1]
3 years ago
8

Write the equation for the line that contains the points (5,6) and (10,12). [No picture this is the full question]

Mathematics
1 answer:
Marina CMI [18]3 years ago
5 0

Answer:

y=1 and 1/5x

Step-by-step explanation:

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E. What is the value of 60 tens?
prohojiy [21]

Answer:

The answer is 600

Step-by-step explanation:

Because 60×10=600.

8 0
2 years ago
Read 2 more answers
Write an exponential function in the form y=ab^ x that goes through the points 007 and 5224​
kati45 [8]
Answer:

y = 19 (3)x

Explanation:

y = abx

Plug in the numbers from the 2 given points to find a and b.

19 = ab0 = a
a = 19

171 = 19b2
b = 3

y = 19 (3)x
3 0
3 years ago
What is the value x² + 3y / xy, if x=4 and y=2?
Kisachek [45]

Answer:

Step-by-step explanation:

x^2+3y/xy

4^2+3*2/4*2

16+6/8

22/8

5 0
3 years ago
The function y = x − 7 x is a solution of the DE xy' + y = 2x. Find x0, given the first-order IVP xy' + y = 2x, y(x0) = 6. Enter
s344n2d4d5 [400]

Answer:

x_0=\{-1,7\}, I_s=(-\infty,0), I_l=(0,+\infty)

Step-by-step explanation:

Remember that an IVP (initial value problem) of first order consists of solving a first-order differential equation (DE) and the solution y(x) must satisfy the <em>initial condition</em> y(x_0)=y_0. This means that the solution y(x) must pass through the point on the plane (x_0,y_0) .

In this case, we know that y(x)=x-\dfrac{7}{x} is already a solution of the DE xy'+y=2x, the value of y_0=6 and we have to find the values x_0 for which the equality y(x_0)=6 holds, that is, y(x_0)=x_0-\dfrac{7}{x_0}=6.

Multiplying by x_0 the above equation we obtain 6x_0=x_0^2-7, where it follows by susbtracting 6x_0 that \\x_0^2-6x_0-7=0. This is a <em>second-degree polynomial equation</em> and its solutions can be found by factoring: x_0^2-6x_0-7=(x_0-7)(x_0+1)=0. Therefore, there are two possible values for x_0: 7 and -1. From here we can see that -1 is the smaller and 7 is the larger.

Now, the larger interval I_s for which y(x) is a solution of the IVP xy'+y=2x, y(-1)=6 (the smaller value of x_0) is the maximal interval where y(x) is defined and pass through the point (-1,6). In this case, it must be that I_s=(-\infty, 0).

Finally, the larger interval I_l for which y(x) is a solution of the IVP xy'+y=2x, y(7)=6 (the larger value of x_0) is the maximal interval where y(x) is defined and pass through the point (7,6). In this case, it must be that I_l=(0,+\infty).

4 0
3 years ago
If B is the midpoint of AC, AC= CD, AB=3x+4, AC= 11x-17, and CE= 49 , find DE
melisa1 [442]

Answer:

Ehhh

Step-by-step explanation:

4 0
3 years ago
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