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3 years ago

When do you use a solid line for graphing a linear inequality?

1 answer:

4 0

Answer: You use it to represent that the equation your graphing is either greater than or equal too or less than or equal too

Step-by-step explanation:

A dotted line is there to represent that the equation your graphing is either grater than or less than

So the solid line represents the same thing, except that they both have a possibility of being equal too which is why the line is solid.

I hope this helps!

You might be interested in

Gils aunt cut his birthday cake into 33 equal pieces eighteen pieces were eaten at his birthday part what fraction of the cake w

kozerog [31]

Answer:

15/33

Step-by-step explanation:

total pieces=33

Pieces eaten=18

Left=33-18

=15

Fraction=15/33

7 0

3 years ago

Solve the initial value problem where y′′+4y′−21 y=0, y(1)=1, y′(1)=0 . Use t as the independent variable.

igor_vitrenko [27]

Answer:

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

$m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7$

So, the solution of the equation is

$y = Ae^{m_{1} t} + Be^{m_{2} t}$

where m₁ = 3 and m₂ = -7.

So,

$y = Ae^{3t} + Be^{-7t}$

Also,

$y' = 3Ae^{3t} - 7e^{-7t}$

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

$y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1 (1)$

$y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}$

Substituting A into (1) above, we have

$\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1 \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}$

Substituting B into A, we have

$A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}$

Substituting A and B into y, we have

$y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

So the solution to the differential equation is

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

6 0

3 years ago

3. What three numbers are the Pythagorean triple generated by using 4 for x and 1 for y. Remember to use the formulas:

Doss [256]

Given:

Consider the three number are given by:

$a=x^2-y^2$

$b=2xy$

$c=x^2+y^2$

To find:

The three numbers are the Pythagorean triple generated by using 4 for x and 1 for y.

Solution:

We have,

$a=x^2-y^2$

$b=2xy$

$c=x^2+y^2$

Substituting 4 for x and 1 for y, we get

$a=4^2-1^2$

$a=16-1$

$a=15$

Similarly,

$b=2(4)(1)$

$b=8$

And,

$c=4^2+1^2$

$c=16+1$

$c=17$

The three three numbers are the Pythagorean triple are 8,15 and 17.

Therefore, the correct option is C.

8 0

2 years ago

Three pieces of artwork were chosen from each grade of Eastlake Elementary School to be displayed at the school art show. If the

kkurt [141]

9×3

=27 pieces of artwork to be displayed.

7 0

2 years ago

Read 2 more answers

cost of baseball tickets can be represented by c(t), where t represents the number of tickets purchased. which correctly shows t

solong [7]

C(t)=30t

Since 120/4= 30

That means it is $30 per ticket

If t represents the number of tickets bought and it’s $30 per ticket it’s 30t to find the total cost

4 0

2 years ago